mechanical decision for a class of integral inequalities

. RESEARCH PAPERS .

SCIENCE CHINAInformation Sciences

September 2010 Vol. 53 No. 9: 1800–1815

doi: 10.1007/s11432-010-4037-2

c© Science China Press and Springer-Verlag Berlin Heidelberg 2010 info.scichina.com www.springerlink.com

Mechanical decision for a class of integral inequalities

YANG Lu1,3, YU WenSheng1,2∗ & YUAN RuYi2

1Shanghai Key Laboratory of Trustworthy Computing, Software Engineering Institute,East China Normal University, Shanghai 200062, China;

2The Key Laboratory of Complex Systems and Intelligence Science, Institute of Automation,Chinese Academy of Sciences, Beijing 100190, China;

3Laboratory for Automated Reasoning and Programming, Chengdu Institute of Computer Applications,Chinese Academy of Sciences, Chengdu 610041, China

Received July 24, 2009; accepted April 30, 2010

Abstract A class of integral inequalities is transformed into homogeneous symmetric polynomial inequalities

beyond Tarski model, where the number of elements of the polynomial, say n, is also a variable and the

coefficients are functions of n. This is closely associated with some open problems formulated recently by Yang

et al. Using Timofte’s dimension-decreasing method for symmetric polynomial inequalities, combined with

the inequality-proving package BOTTEMA and a program of implementing the method known as successive

difference substitution, we provide a procedure for deciding the nonnegativity of the corresponding polynomial

inequality such that the original integral inequality is mechanically decidable; otherwise, a counterexample will

be given. The effectiveness of the algorithm is illustrated by some more examples.

Keywords integral inequality, symmetric polynomial inequality, Timofte’s dimension-decreasing method,

successive difference substitution, mechanical decision, inequality-proving package BOTTEMA

Citation Yang L, Yu W S, Yuan R Y. Mechanical decision for a class of integral inequalities. Sci China Inf Sci,

2010, 53: 1800–1815, doi: 10.1007/s11432-010-4037-2

1 Introduction

The mechanical decision for polynomial inequalities is always a hard and hot topic in mathematicsmechanization and automated reasoning. A great progress in this area has been made in recent years (see[1] for some details). Many “moment inequalities” [2–6] appear in random analysis and mechanics analysis,which are actually homogeneous integral inequalities. However, there are few studies on mechanicaldecision for integral inequalities.

This paper is concerned with the mechanical decision for a class of integral inequalities which can betransformed into homogeneous symmetric polynomial inequalities. Using Timofte’s dimension-decreasingmethod for symmetric polynomial inequalities [7–9], combined with the inequality-proving package BOT-TEMA [1, 10–12] and a program which implements the method known as successive difference substitution[1, 13–17], we provide a procedure to decide the nonnegativity of the corresponding polynomial inequalitysuch that the original integral inequality is mechanically decidable. For simplicity, all integrals mentionedin this paper are assumed to be Riemann’s [18]. First, Let us take an example below.

∗Corresponding author (email: [email protected])

YANG Lu, et al. Sci China Inf Sci September 2010 Vol. 53 No. 9 1801

Example 1. Does the following integral inequality (1) hold for any function g(s) which is integrableon [0, 1]?

G =∫ 1

0

|g(s)|7ds

∫ 1

0

|g(s)|ds − 3∫ 1

0

g6(s)ds

∫ 1

0

g2(s)ds

+ 3∫ 1

0

|g(s)|5ds

∫ 1

0

|g(s)|3ds −( ∫ 1

0

g4(s)ds

)2

� 0. (1)

According to the definition of Riemann integral [18], we divide the interval [0, 1] into n parts

P : 0 = s0 < s1 < · · · < sn = 1.

Obviously, ∫ 1

0

g(s)ds = limn→∞

n∑i=1

1n

g

(i

n

).

It follows that

G = limn→∞

(( n∑i=1

1n

∣∣∣∣g(

i

n

)∣∣∣∣7)( n∑

i=1

1n

∣∣∣∣g(

i

n

)∣∣∣∣)− 3

( n∑i=1

1n

g6

(i

n

))( n∑i=1

1n

g2

(i

n

))

+ 3( n∑

i=1

1n

∣∣∣∣g(

i

n

)∣∣∣∣5)( n∑

i=1

1n

∣∣∣∣g(

i

n

)∣∣∣∣3)

−( n∑

i=1

1n

g4

(i

n

))2).

Proving G � 0 is equivalent to verifying that the following symmetric polynomial inequality of degree8 in n variables holds for all positive integers n and arbitrary nonnegative variables xi:

f0 =n∑

i=1

x7i

n

n∑i=1

xi

n− 3

n∑i=1

x6i

n

n∑i=1

x2i

n+ 3

n∑i=1

x5i

n

n∑i=1

x3i

n−

( n∑i=1

x4i

n

)2

� 0. (2)

Thus, the decision for original integral inequality can be transformed into that for a homogeneoussymmetric polynomial inequality f0 � 0. Since n itself is a variable, the corresponding homogeneoussymmetric polynomial inequality is beyond typical Tarski model.

In the beginning of 1950s, Tarski [19] published his famous paper: “A decision method for elemen-tary for Algebra and Geometry”; he proved that any proposition in elementary algebra and elementarygeometry can be decided mechanically. This kind of problems belongs to Tarski model which always ismechanically decidable. Any determined propositional sentence in Tarski model has a specified numberof algebraic variables. According to the well-known Godel’s incompleteness theorem, the mechanicallydecidable problems are rare in mathematics. Even in elementary number theory, the sentences are notmechanically decidable in general. Even so, research on automated theorem proving has been flourishingfor the recent 30 years, a variety of new methods have been put forward and new results have beenachieved one after another in this area such as Wu’s method [20–23], Grobner basis method [24–26], re-sultant method [27], numerical parallel method [28], point-eliminating method that can produce readableproofs [29], cylindrical algebraic decomposition [30–34], etc. The problems that can be dealt with bythese methods almost all belong to Tarski model.

Some inequalities that come from informatics, technological and mathematical sciences may dependupon a discrete parameter n. These problems are beyond Taraki model, e.g. inequalities with uncertainnumber of variables or those proven by traditional induction. Generally speaking, the propositions withindefinite number of variables are not mechanically decidable, but some of them may be transformedinto the mechanically decidable type. Refs. [35, 36] simulated induction by computer and proved manyinequalities, such as Cauch-Schwarz inequality, although that is not a decision algorithm, and ref. [1] didthe same thing by using an inequality proving package BOTTEMA. A recent paper [37], published inScience in China, also gave a practical algorithm to decide the nonnegativity of a class of polynomialswith uncertain number of elements. A program “nprove” implemented in Maple platform can do thisefficiently [1, 37].

1802 YANG Lu, et al. Sci China Inf Sci September 2010 Vol. 53 No. 9

In [37], a class of symmetric polynomial inequalities with degree less than 5 is successfully transformedinto finitely many inequalities independent of n by using Timfote dimension-decreasing algorithm [7–9]. The nonnegativity of these polynomials can be decided by using the inequality-proving packageBOTTEMA [1, 10–12]. How to generalize the result of [37] to deal with the polynomials with degreegreater than 5 or coefficients depending on n? This is a challenging problem.

In this paper, a class of integral inequalities is transformed into homogeneous symmetric polynomialinequalities beyond Tarski model, where the number of elements of the polynomial, say n, is also avariable and the coefficients are functions of n. This is closely associated with some open problemsformulated recently by Yang et al. [37]. Using Timofte’s dimension-decreasing method for symmetricpolynomial inequalities [7–9], combined with the inequality-proving package BOTTEMA [1, 10–12] anda program of implementing successive difference substitution [1, 13–17], we provide a procedure to decidethe nonnegativity of the corresponding polynomial inequality, hence the original integral inequality ismechanically decidable. If it does not hold, a counterexample will be given. The effectiveness of thealgorithm is illustrated by some more examples.

The rest of the paper is organized as follows. In section 2, we review Timofte dimension-decreasingapproach [7–9], describe the mechanically decidable problem for a class of symmetric polynomial inequal-ities and present a complete theoretical result. In section 3, we improve the efficiency of the algorithmby employing successive difference substitution [1, 13–17], and present a further discussion on alternativeapproaches to dimension-decreasing. In section 4, an algorithm is described in detail and some examplesare demonstrated to show the effectiveness. Finally, the conclusions are contained in section 5.

2 Timofte’s dimension-decreasing approach and mechanical decision for non-negativity of polynomials

In this section, we review the Timefote’s dimension-decreasing approach [7–9] and describe the mechan-ically decidable problems for a class of symmetric polynomial inequalities. A complete theoretical resultis given. Similar to [1, 37], the definitions and notations are as follows.

Definition 1. A homogeneous polynomial f(x1, x2, . . . , xn) is called symmetric if it holds that

f(x1, x2, . . . , xn) = f(σ(x1, x2, . . . , xn))

for any σ ∈ Sn, where Sn is a full permutation group of n objects.A homogeneous polynomial is called a form for short. Let Rn denote the n-dimensional real number

space, and Rn+ the nonnegative subspace of Rn. By Sn,m denote the set of m-degree symmetric form on

Rn, which is a vector space under the usual addition and scale multiplication. By dim(Sn,m) denote thedimension of Sn,m.

Definition 2. For each positive integer k and a point (x1, x2, . . . , xn) ∈ Rn, the kth symmetric powersum is defined as

pk(x) =n∑

i=1

xki .

Lemma 1 (The fundamental theorem on symmetric form)[1, 37]. Any symmetric form f ∈ Sn,m can beexpressed as a unique polynomial of (p1, p2, . . . , pn), where d = min(n, m), and p1, p2, . . . , pn are algebraicindependent, i.e., there is no nonzero polynomial g such that g(p1, p2, . . . , pn) = 0.

By Ω denote the set of nonnegative integer solutions to the indefinite equation

1y1 + 2y2 + · · · + dyd = m.

Lemma 2 [1, 37]. The set Bn,m = {pλ11 pλ2

2 · · · pλd

d |(λ1, λ2, . . . , λd) ∈ Ω} is a basis of vector space Sn,m.The dimension of Sn,m is the cardinal number of Ω, denoted by |Ω|, here d = min(n, m).


According to Lemma 2, homogeneous symmetric polynomial of degree 5 in n variables f ∈ Sn,5 canbe expressed as

f = a1p5 + a2p4p1 + a3p3p2 + a4p3p21 + a5p

22 + a6p2p

31 + a7p

51

:= ([a1, a2, . . . , a7]p).

Homogeneous symmetric polynomial of degree 6 in n variables f ∈ Sn,6 can be expressed as

f = a1p6 + a2p5p1 + a3p4p2 + a4p4p21 + a5p

23 + a6p3p2p1 + a7p3p

31 + a8p

32 + a9p

22p

21 + a10p2p

41 + a11p

61

:= ([a1, a2, . . . , a11]p).

Furthermore, homogeneous symmetric polynomial of degree m in n variables f ∈ Sn,m can be expressedas

f = a1pm + a2pm−1p1 + a3pm−2p2 + a4pm−2p21 + · · · + a|Ω|pm

1

:= ([a1, a2, . . . , a|Ω|]p),

where |Ω| stands for the cardinal number of set Ω.Some notations are introduced:

1k = (1, 1, . . . , 1︸︷︷︸)k

;0k = (0, 0, . . . , 0︸︷︷︸)

k.

The notation [z] stands for the largest integral part that is smaller than or equal to the real numberz ∈ R. For every x = (x1, x2, . . . , xn) ∈ Rn, we denote

v(x) = |{xj |j = 1, 2, . . . , n}|,v∗(x) = |{xj |xj �= 0, j = 1, 2, . . . , n}|,

where |A| stands for cardinal number of set A. Intuitively speaking, v(x) counts the distinct componetsof x, while v∗ counts the non-zero distinct componets only.

Lemma 3 [1, 7–9, 37]. (i) A symmetric polynomial inequality of degree m holds on Rn+, if and only if

it holds on set {x|x ∈ Rn+, v∗(x) � max ([m

2 ], 1)}; (ii) A symmetric inequality of degree m holds in Rn, ifand only if it holds in {x|x ∈ Rn, v(x) � max ([m

2 ], 2)}.Lemma 4. For any f ∈ Sn,m, the following equivalence relation holds:

f(x) � 0, x ∈ Rn+ ⇐⇒ f(t1 · 1r1, t2 · 1r2, . . . , tl · 1rl

, 0n−r1−r2−···−rl) � 0,

where l = max ([m2 ], 1), ∀t = (t1, t2, . . . , tl) ∈ Rl

+, ∀r = (r1, r2, . . . , rl) ∈ Nl,n, Nl,n := {(r1, r2, . . . , rl)|r1, r2, . . . , rl are positive integers and r1 + r2 + · · · + rl � n}.Proof. From Lemma 3, it follows that

f(x) � 0, x ∈ Rn+ ⇐⇒ f(y) � 0, ∀y ∈ Rn

+, v∗(y) � l = max([

m

2

], 1

).

Since f is symmetric, let y = (t1 ·1r1, t2 ·1r2 , . . . , tl ·1rl, 0n−r1−r2−···−rl

), ∀t = (t1, t2, . . . , tl) ∈ Rl+, ∀r =

(r1, r2, . . . , rl) ∈ Nl,n, Nl,n = {(r1, r2, . . . , rl)|r1, r2, . . . , rl be positive integers and r1 + r2 + · · ·+ rl � n}.This completes the proof.

For the sake of convenience, for any f ∈ Sn,m, l = max ([m2 ], 1), and ∀t = (t1, t2, . . . , tl) ∈ Rl

+, ∀r =(r1, r2, . . . , rl) ∈ Nl,n, Nl,n = {(r1, r2, . . . , rl)|r1, r2, . . . , rl are positive integers and r1 + r2 + · · ·+ rl � n},denote

qk =l∑

i=1

ritki , k = 1, 2, . . . , m.


For any f ∈ Sn,m, denote

fr(t) = a1qm + a2qm−1q1 + a3qm−2q2 + a4qm−2q21 + · · · + a|Ω|qm

1

:= ([a1, a2, . . . , a|Ω|]q(r,t)).

From Lemma 4, it is straightforward to get

Lemma 5. Suppose that f(x1, x2, . . . , xn) ∈ Sn,m, (x1, x2, . . . , xn) ∈ Rn+. Then f(x1, x2, . . . , xn) � 0

holds for any positive integer n if and only if fr(t) � 0 holds for any nonnegative integers r1, r2, . . . , rl,and ∀t = (t1, t2, . . . , tl) ∈ Rl

+, where l = max ([m2 ], 1).

According to Lemma 5, deciding the nonnegativity of a symmetric polynomial f ∈ Sn,m can betransformed into deciding the inequality fr(t) � 0. Since r is indefinite, this problem is beyond Tarskimodel and cannot be solved by computer. So we have to do more things to transform the problem intosome Tarski’s statements.

As we know, the integral inequality can be transformed into homogeneous symmetric polynomialinequality beyond Tarski model. The corresponding polynomial can be written as a new polynomialwhose every element is an average power sum of the original variables. The property of this class ofsymmetric polynomials is explored in the following. Some definitions and notations are given below.

Definition 3. For any positive integer k and a fix point (x1, x2, . . . , xn) ∈ Rn, the average of kth powersum is defined as

Ak(x) =∑n

i=1 xki

n.

In what follows, we focus on symmetric form of degree m in n variables that can be expressed aspolynomial of (A1, A2, . . . , Ad) in rational field, where d := min(n, m). Similarly, let Ω denote nonnegativeinteger solutions of indefinite equation

1y1 + 2y2 + · · · + dyd = m,

|Ω| stands for the cardinal number of set Ω. For the sake of convenience, the set of all such symmetricform of degree m in n variables is denoted by S0

n,m. It can be seen that the symmetric forms are quitegeneral.

The symmetric homogeneous polynomial of degree 5 in n variables f0 ∈ S0n,5 can be expressed as

f0 = a1A5 + a2A4A1 + a3A3A2 + a4A3A21 + a5A

22 + a6A2A

31 + a7A

51

:= ([a1, a2, . . . , a7]A),

and the symmetric homogeneous polynomial of degree 6 in n variables f0 ∈ S0n,6 can be expressed as

f0 = a1A6 + a2A5A1 + a3A4A2 + a4A4A21 + a5A

23 + a6A3A2A1 + a7A3A

31

+ a8A32 + a9A

22A

21 + a10A2A

41 + a11A

61

:= ([a1, a2, . . . , a11]A).

Furthermore, the symmetric homogeneous polynomial of degree m in n variables f0 ∈ S0n,m can be

expressed as

f0 = a1Am + a2Am−1A1 + a3Am−2A2 + a4Am−2A21 + · · · + a|Ω|Am

1

:= ([a1, a2, . . . , a|Ω|]A).

The problem investigated here is

Question. Given a homogenous symmetric polynomial of degree m in n variables f0(x1, x2, . . . , xn) ∈S0

n,m with rational coefficients, is the following proposition mechanically decidable?

Inequality f0(x1, x2, . . . , xn) � 0 holds for any positive integer n and any xi ∈ R+.


Obviously, the nonnegativity of f0 = [a1, a2, a3, a4, a5, a6, a7]A ∈ S0n,5 is equivalent to the nonnegativity

of f = [n4a1, n3a2, n3a3, n2a4, n3a5, na6, a7]p ∈ Sn,5. The nonnegativity of f0 = [a1, a2, a3, a4, a5, a6,

a7, a8, a9, a10, a11]A ∈ S0n,6 is equivalent to the nonnegativity of f = [n5a1, n4a2, n4a3, n3a4, n4a5, n3a6,

n2a7, n3a8, n2a9, na10, a11]p ∈ Sn,6. Generally speaking, the nonnegativity of f0 = [a1, a2, a3, a4, . . . ,

a|Ω|]A ∈ S0n,m is equivalent to the nonnegativity of f = [nm−1a1, nm−2a2, nm−2a3, nm−3a4, . . . , a|Ω|]p ∈

Sn,m.For any f0 ∈ S0

n,m, l = max ([m2 ], 1), and ∀t = (t1, t2, . . . , tl) ∈ Rl

+, ∀r = (r1, r2, . . . , rl) ∈ Nl,n, Nl,n ={(r1, r2, . . . , rl)|r1, r2, . . . , rl are nonnegative integers and r1 + r2 + · · · + rl � n}. Denote

Bk =∑l

i=1 ritki

n, k = 1, 2, . . . , m.

For any f0 ∈ S0n,m, denote

f0r (t) := a1Bm + a2Bm−1B1 + a3Bm−2B2 + a4Bm−2B

21 + · · · + a|Ω|Bm

1

:= ([a1, a2, . . . , a|Ω|]B(r,t)),

that is,

f0r (t) = a1

(r1tm1 + r2t

m2 + · · · + rlt

ml )

n

+ a2

(r1t

m−11 + r2t

m−12 + · · · + rlt

m−1l

)n

(r1t1 + r2t2 + · · · + rltl)n

+ a3

(r1t

m−21 + r2t

m−22 + · · · + rlt

m−2l

)n

(r1t

21 + r2t

22 + · · · + rlt

2l

)n

+ a4

(r1t

m−21 + r2t

m−22 + · · · + rlt

m−2l

)n

(r1t1 + r2t2 + · · · + rltl)n2

2

+ · · · + a|Ω|(r1t1 + r2t2 + · · · + rltl)

nm

m

.

Denote

f0r (t) := a1 (r1t

m1 + r2t

m2 + · · · + rlt

ml )

+ a2

(r1t

m−11 + r2t

m−12 + · · · + rlt

m−1l

)(r1t1 + r2t2 + · · · + rltl)

+ a3

(r1t

m−21 + r2t

m−22 + · · · + rlt

m−2l

) (r1t

21 + r2t

22 + · · · + rlt

2l

)+ a4

(r1t

m−21 + r2t

m−22 + · · · + rlt

m−2l

)(r1t1 + r2t2 + · · · + rltl)

2

+ · · · + a|Ω| (r1t1 + r2t2 + · · · + rltl)m

:= ([a1, a2, · · · , a|Ω|]q(r,t)).

From Lemma 4, it follows that

Lemma 6. Suppose that f0(x1, x2, . . . , xn) ∈ S0n,m, (x1, x2, . . . , xn) ∈ Rn

+. Then f0(x1, x2 , . . . , xn)� 0 holds for all nonnegative integers n if and only if f0

r (t) � 0 holds for all nonnegative integers r1, r2,

. . . , rl, n, r1+r2+. . .+rl � n, and ∀t = (t1, t2, . . . , tl) ∈ Rl+; or, equivalently, if and only if f0

r (t) � 0 holdsfor any nonnegative rational numbers r1, r2, . . . , rl, r1 +r2+ · · · + rl � 1, and ∀t = (t1, t2, . . . , tl) ∈ Rl

+;or, equivalently, if and only if f0

r (t) � 0 holds for any nonnegative real numbers r1, r2, . . . , rl, r1 +r2

+ . . . +rl � 1, and ∀t = (t1, t2, . . . , tl) ∈ Rl+, here l = max ([m

2 ], 1).

Proof. The first equivalent condition of Lemma 6 can be directly obtained from Lemma 4.The second equivalent condition can be obtained from the arbitrariness of the nature numbers r1, r2,

. . . , rl, n.To prove the third equivalent condition, assume that there exist nonnegative real numbers rR1, rR2,

. . . , rRl, rR1 + rR2+ · · · + rRl � 1, and t = (t1, t2, . . . , tl) ∈ Rl+, such that f0

rR(t) < 0, where l =

max ([m2 ], 1). Because of the continuity of function f0

r (t), there must exist nonnegative rational numbers


rQ1, rQ2, . . . , rQl, rQ1 + rQ2 + · · · + rQl � 1, and t = (t1, t2, . . . , tl) ∈ Rl+, such that f0

rQ(t) < 0. This

contradicts the second equivalent condition. This completes the proof.By the third equivalent condition of Lemma 6, we can obtain the main result result of the paper.

Theorem 1. Given a homogenous symmetric polynomial of degree m in n variables f0(x1, x2, . . . , xn)∈ S0

n,m with rational coefficients, the following proposition is mechanically decidable:

Inequality f0(x1, x2, · · · , xn) � 0 holds for any positive integer n and all xi ∈ R+.

By Theorem 1 and Lemma 6, deciding the the nonnegativity of a symmetric polynomial f0 ∈ S0n,m

has been transformed into deciding the inequality f0r (t) � 0 with finite variables in constraints of r =

(r1, r2, . . . , rl) ∈ Rl+, t = (t1, t2, . . . , tl) ∈ Rl

+ and r1 + r2 + · · · + rl < 1. Deciding the inequalityf0

r (t) � 0 obviously belongs to Tarski model [1, 19, 21–23, 27, 37]. Then it can be carried out by the celldecomposition [30–34], which is often used to prove algebraic inequalities. There are a few softwares fordealing with cell decomposition such as DISCOVERER [1, 38, 39] and BOTTEMA [1, 10–12]. Amongthem, BOTTEMA [1, 10–12] appears preferably for our purpose. Theoretically, it suffices to use the“xprove” command in BOTTEMA [1, 10–12].

3 Further discussion on dimension-decreasing approach

In the above section, deciding the nonnegativity of a symmetric form of degree m in n variables hasbeen successfully transformed into deciding an inequality with some constraints, belonging to Tarskimodel [1, 19, 21–23, 27, 37]. The cell decomposition algorithm [30–34] can be applied to deciding thenonnegativity of polynomials. However, since the cell decomposition algorithm has high complexity, itcan only deal with polynomials only in few variables with lower degrees in practice. In what follows,successive difference substitution method [1, 13–15] is applied to reduce the computational complexity.It is noticed that successive difference substitution [1, 13–15] with lesser mathematics and without celldecomposition is efficient for our problems. More recently, refs. [16, 17] discussed the termination ofsuccessive difference substitution and showed that this method is applicable to a very extensive class ofpolynomials.

For any f0 ∈ S0n,m, consider

f0r (t) = a1

(r1tm1 + r2t

m2 + · · · + rlt

ml )

n

+ a2(r1t

m−11 + r2t

m−12 + · · · + rlt

m−1l )

n

(r1t1 + r2t2 + · · · + rltl)n

+ a3(r1t

m−21 + r2t

m−22 + · · · + rlt

m−2l )

n

(r1t21 + r2t

22 + · · · + rlt

2l )

n

+ a4(r1t

m−21 + r2t

m−22 + · · · + rlt

m−2l )

n

(r1t1 + r2t2 + · · · + rltl)n2

2

+ · · · + a|Ω|(r1t1 + r2t2 + · · · + rltl)

nm

m

.

Let rl+1 := n − r1 − r2 − · · · − rl. Denote

F 0r (t) :=

( l+1∑i=1

ri

)m−1

a1(r1tm1 + r2t

m2 + · · · + rlt

ml )

+( l+1∑

i=1

ri

)m−2

a2(r1tm−11 + r2t

m−12 + · · · + rlt

m−1l )(r1t1 + r2t2 + · · · + rltl)

+( l+1∑

i=1

ri

)m−2

a3(r1tm−21 + r2t

m−22 + · · · + rlt

m−2l )(r1t

21 + r2t

22 + · · · + rlt

2l )


+( l+1∑

i=1

ri

)m−3

a4(r1tm−21 + r2t

m−22 + · · · + rlt

m−2l )(r1t1 + r2t2 + · · · + rltl)2

+ · · · + a|Ω|(r1t1 + r2t2 + · · · + rltl)m

=([( l+1∑

i=1

ri

)m−1

a1,

( l+1∑i=1

ri

)m−2

a2, . . . , a|Ω|

]q(r,t)

).

From Lemma 6, it follows that:


+. Then f0(x1, x2, . . . , xn) �0 holds for any nature number n if and only if F 0

r (t) � 0 holds for any nonnegative real numberr1, r2, . . . , rl, rl+1, and ∀t = (t1, t2, . . . , tl) ∈ Rl

+, where l = max ([m2 ], 1).

The inequality to be verified in Lemma 7 is free of constraints, and can be easily dealt with by successivedifference substitution.

By now, using Timofte dimension-decreasing approach [1, 7–9, 37], the nonnegativity decision problemfor symmetric form of degree m in n variables in S0

n,m has been successfully transformed into a problemof verifying inequalities in 2l (or 2l + 1) variables, where l depends on the degrees of given symmetricform. l may be very large when the degree of given symmetric form is high.

An alternative dimension-decreasing approach, which does not depend on the degree of given symmetricform and is significantly different from Timofte dimension-decreasing approach, has been used effectivelyin [40, 41]. We will use a similar approach to deal with our problems. The corresponding inequalitiesobtained by this approach are also free of constraints, and can be directly verified by successive differencesubstitution. The variables included in these inequalities may be less than 2l sometimes.

Consider homogeneous symmetric polynomial f0 ∈ S0n,m of degree m in n variables,

f0 = a1Am + a2Am−1A1 + a3Am−2A2 + a4Am−2A21 + · · · + a|Ω|Am

1 .

In order to decide the nonnegativity of f0(x1, x2, . . . , xn) over the nonnegative real number space Rn+,

we calculate its stationary point equations

df0

dxi= a1

dAm

dxi+ a2

dAm−1A1

dxi+ a3

dAm−2A2

dxi+ a4

dAm−2A21

dxi+ · · · + a|Ω|

dAm1

dxi= 0, i = 1, 2, . . . , n.

Obviously, df0

dxiis of degree m − 1 at most in xi, and the above stationary point equation has no more

than m − 1 positive real roots about xi. By using the famous Descartes’ law of signs [1, 27] or completediscrimination system [27, 39] for polynomials, an estimation to the number of positive real roots of aboveequation can be obtained easily. In next section, some examples are provided to show the effectivenessof this approach.

Since f0(x1, x2, . . . , xn) ∈ S0n,m is symmetric, the stationary point equation set has the same form for

every variables xi, i ∈ {1, 2, . . . , n}. Assume that the above stationary point equation at most has L

positive real roots. Introduce L nonnegative variables {t1, t2, . . . , tL}. Obviously, if f0(x1, x2, . . . , xn) isnonnegative over the space Rn

+, then f0(x1, x2, . . . , xn) is nonnegative over set {x|xi ∈ {t1, t2, . . . , tL}, i =1, 2, . . . , n}. Conversely, if f0(x1, x2, . . . , xn) is nonnegative when all of its valuables are constrained in{t1, t2, . . . , tL} such that f0(x1, x2, . . . , xn) are nonnegative at all possible stationary points, then f0 isnonnegative.

More specifically, denote

qk =L∑

i=1

ritki , k = 1, 2, . . . , m.

Like Lemma 7, denote

F 0r (t) :=

( L∑i=1

ri

)m−1

a1(r1tm1 + r2t

m2 + · · · + rLtmL )


+( L∑

i=1

ri

)m−2

a2(r1tm−11 + r2t

m−12 + · · · + rLtm−1

L )(r1t1 + r2t2 + · · · + rLtL)

+( L∑

i=1

ri

)m−2

a3(r1tm−21 + r2t

m−22 + · · · + rLtm−2

L )(r1t21 + r2t

22 + · · · + rLt2L)

+( L∑

i=1

ri

)m−3

a4(r1tm−21 + r2t

m−22 + · · · + rLtm−2

L )(r1t1 + r2t2 + · · · + rLtL)2

+ · · · + a|Ω|(r1t1 + r2t2 + · · · + rLtL)m

:=([( L∑

i=1

ri

)m−1

a1,

( L∑i=1

ri

)m−2

a2, · · · , a|Ω|

]q(r,t)

).

Similar to the proof of Lemma 6, we have


+. Then f0(x1, x2, . . . , xn) � 0holds for all nonnegative integers n if and only if F 0

r (t) � 0 holds for all nonnegative integers r1, r2, . . . , rL,

and ∀t = (t1, t2, . . . , tL) ∈ RL+; or, equivalently, if and only if F 0

r (t) � 0 holds for any nonnegative rationalnumbers r1, r2, . . . , rL, and ∀t = (t1, t2, . . . , tL) ∈ RL

+; or, equivalently, if and only if F 0r (t) � 0 holds for

any nonnegative real numbers r1, r2, . . . , rL, and ∀t = (t1, t2, . . . , tL) ∈ RL+, where L � m−1 is the upper

bound of number of nonnegative real roots of the stationary point equations set df0

dxi= 0, i = 1, 2, . . . , n.

Proof. Consider the following stationary point equations:

df0

dxi= 0, i = 1, 2, . . . , n.

Since f0(x1, x2, . . . , xn) ∈ S0n,m is symmetric, the stationary point equation set has the same form for

every variables xi, i ∈ {1, 2, . . . , n}. Assume that the above stationary point equation at most has L

positive real roots. Introduce L nonnegative variables {t1, t2, . . . , tL}.The first equivalent condition can be directly obtained from the above discussion.The second equivalent condition can be obtained from the arbitrariness of the nature numbers r1, r2, . . . ,

rL.

In order to prove the third equivalent condition, assume that there exist nonnegative real numbers rR1,

rR2, . . . , rRL, t = (t1, t2, . . . , tL) ∈ RL+, so that F 0

r (t) < 0 holds. Because of the continuity of functionF 0

r (t), there must exist nonnegative rational numbers rQ1, rQ2, . . . , rQL, and t = (t1, t2, . . . , tL) ∈ RL+,

such that F 0r (t) < 0. This contradicts the second condition. This completes the proof.

According to Lemma 8, the nonnegativity decision problem of symmetric form S0n,m of degree m in

n variables has been successfully transformed into the problem of verifying inequalities in 2L variables.These inequalities are free of constraints, and can be verified by successive difference substation methoddirectly. The number of variables 2L may be less than 2l sometimes.

In the end of this section, it is worth pointing out that some techniques can be used to determine thenonnegativity of F 0

r (t) by investigating its special structure without successive difference substitution.More details can be found in Remark 1 and Remark 2 in the following section. Successive differencesubstitution is a readable [1] mechanical decision method, whereas the method for getting the nonnega-tivity of F 0

r (t) directly from its expression, without successive difference substitution, can be viewed as acertificate [1] mechanical decision method.

4 Algorithm design and examples

Based on the discussion in the previous two sections, according to Theorem 1 and Lemmas 6–8, analgorithm “nm0prove” can be given as follows:


Algorithm (nm0prove)Input: A homogeneous symmetric polynomial with rational coefficients of degree m in n variables

f0(x1, x2, . . . , xn) ∈ S0n,m.

Output: Return “true” if inequality f0(x1, x2, . . . , xn) � 0 holds for all positive integers n and all(x1, x2, . . . , xn) ∈ Rn

+; return “false” otherwise.N1: Compute l and L separately, i.e., determine the number of variables in the inequality to be

verified.N2: Compute f0

r (t), F 0r (t) and F 0

r (t).N3: Call Maple command “collect(·, distributed)” and “factor” to rearrange f0

r (t), F 0r (t) and F 0

r (t)respectively. Check the non-negativeness of each coefficients. If we can make sure that f0

r (t) � 0, orF 0

r (t) � 0, or F 0r (t) � 0, then return “true” and stop, otherwise go to next step.

N4: Use software package BOTTEMA or successive difference substitution to determine the nonnega-tivity of f0

r (t) (with constraints), F 0r (t) and F 0

r (t). If we can make sure that f0r (t) � 0 (with constraints),

or F 0r (t) � 0, or F 0

r (t) � 0, then return “true” and stop; otherwise return “false” and stop.

It should be noticed that the commands in each step of the above algorithm can run in parallel. Wecan choose more efficient commands according to the specific examples.

Some examples are given in the following. All results are obtained from Maple platform on a Pentiumpersonal desktop computer with dual CPU (memory 2.0 GB).

First, consider the example mentioned in section 1.

Example 1. In section 1, we have already equivalently transformed the decision problem for homoge-neous integral inequalities (1) into the decision problem of average power sum inequalities (2).

According to Lemma 6, we need to prove the following inequality:

f0r =

4∑i=1

rit7i

4∑i=1

riti − 34∑

i=1

rit6i

4∑i=1

rit2i + 3

4∑i=1

rit5i

4∑i=1

rit3i −

( 4∑i=1

rit4i

)2

� 0,

where t = (t1, t2, t3, t4) ∈ R4+, r1, r2, r3, r4 are arbitrary nonnegative real numbers, satisfying r1 + r2 +

r3 + r4 � 1.

By Lemma 7, we need to prove the following inequality:

F 0r =

( 5∑i=1

ri

)6 4∑i=1

rit7i

4∑i=1

riti − 3( 5∑

i=1

ri

)6 4∑i=1

rit6i

4∑i=1

rit2i

+ 3( 5∑

i=1

ri

)6 4∑i=1

rit5i

4∑i=1

rit3i −

( 5∑i=1

ri

)6( 4∑i=1

rit4i

)2

� 0,

where ∀t = (t1, t2, t3, t4) ∈ R4+, r1, r2, r3, r4, r5 are arbitrary nonnegative real numbers.

Removing the positive factor (∑5

i=1 ri)6 of F 0r , difference substitution program “tsds” is called, and it

returns “true” after running 1103 s. If we call partial difference substitution program, and enter command“sds(F 0

r , {x1, x2, x3, x4})”, then it returns “true” in less than 1 s. Thus for all positive integers n, and∀xi ∈ R+(i = 1, . . . , n), f0 � 0 holds.

According to Lemma 8, compute stationary point equations of f0

df0

dxi= 7A1x

6i − 18A2x

5i + 15A3x

4i − 8A4x

3i + 9A5x

2i − 6A6xi + A7 = 0, i = 1, . . . , n.

Based on Descartes’ law of signs, the above stationary point equations set at most has 6 positive realroots, and we only need to prove the following inequality:

F 0r =

( 6∑i=1

ri

)6 6∑i=1

rit7i

6∑i=1

riti − 3( 6∑

i=1

ri

)6 6∑i=1

rit6i

6∑i=1

rit2i

+ 3( 6∑

i=1

ri

)6 6∑i=1

rit5i

6∑i=1

rit3i −

( 6∑i=1

ri

)6( 6∑i=1

rit4i

)2

� 0,


where ∀t = (t1, . . . , t6) ∈ R6+, ri, i = 1, . . . , 6, are arbitrary nonnegative real numbers.

Removing the positive factor (∑6

i=1 ri)6 of F 0r , partial difference substitution program is called. Enter

the command “sds(F 0r , {x1, x2, x3, x4, x5, x6})”. Then it returns “true” after running 145 s. Thus for all

positive integers n, and ∀xi ∈ R+(i = 1, . . . , n), f0 � 0 holds.

Remark 1. In order to reduce the computational complexity, we can remove both the positive factor(∑5

i=1 ri)6 of F 0r and the positive factor (

∑6i=1 ri)6 of F 0

r .

Remark 2. Sort F 0r (after removing the positive factor (

∑6i=1 ri)6) as variables r1, . . . , r6, by call-

ing Maple command “collect(F 0r , [r1, . . . , r6], distributed)”. The following expression can be found by

factorizing all coefficients of F 0r with respect to variables r1, . . . , r6:

F 0r = G1r1r2 + G2r1r3 + G3r1r4 + G4r1r5 + G5r6r1 + G6r2r3 + G7r2r4 + G8r2r5

+ G9r6r2 + G10r3r4 + G11r3r5 + G12r6r3 + G13r4r5 + G14r6r4 + G15r6r5,

where

G1 = t1t2(t21 + t1t2 + t22)(t1 − t2)4, G2 = t1t3(t21 + t1t3 + t23)(t1 − t3)4,







G15 = t5t6(t25 + t5t6 + t26)(t5 − t6)4.

Since Gi � 0, i = 1, . . . , 15, it is obvious that F 06 � 0. This technique can also be used to F 0

r .

Remark 3. The constraint that the integrable interval should be [0, 1] is not an essential restriction.This method can be extended to integral inequalities with arbitrary integrable interval.

Remark 4. The decision of F 0r � 0 in this example is also considered in [15] using successive difference

substitution program.

Example 2. Assume that g(s) is integrable on interval [0, 1], consider the following integral inequality:

G = 4∫ 1

0

|g(s)|5ds −∫ 1

0

g4(s)ds

∫ 1

0

|g(s)|ds − 4∫ 1

0

|g(s)|3ds

∫ 1

0

g2(s)ds

+ 2∫ 1

0

|g(s)|3ds

( ∫ 1

0

|g(s)|ds

)2

− 3( ∫ 1

0

g2(s)ds

)2 ∫ 1

0

|g(s)|ds

+ 3∫ 1

0

g2(s)ds

( ∫ 1

0

|g(s)|ds

)3

−( ∫ 1

0

|g(s)|ds

)5

� 0.

Does the inequality G � 0 hold for any g(s) which is integrable on [0, 1]? Equivalently, consider thefollowing polynomial inequality:

f0 = 4n∑

i=1

x5i

n−

n∑i=1

x4i

n

n∑i=1

xi

n− 4

n∑i=1

x3i

n

n∑i=1

x2i

n+ 2

n∑i=1

x3i

n

( n∑i=1

xi

n

)2

− 3( n∑

i=1

x2i

n

)2 n∑i=1

xi

n+ 3

n∑i=1

x2i

n

( n∑i=1

xi

n

)3

−( n∑

i=1

xi

n

)5

.

Does f0 � 0 hold for all positive integers n and all xi ∈ R+, i = 1, . . . , n?By Lemma 6, we need to prove the follow inequality:

f0r = 4(r1t

51 + r2t

52) − (r1t

41 + r2t

42)(r1t1 + r2t2) − 4(r1t

31 + r2t

32)(r1t

21 + r2t

22)


+ 2(r1t31 + r2t

32)(r1t1 + r2t2)2 − 3(r1t

21 + r2t

22)

2(r1t1 + r2t2)

+ 3(r1t21 + r2t

22)(r1t1 + r2t2)3 − (r1t1 + r2t2)5 � 0,

where t = (t1, t2) ∈ R2+, r1, r2 are arbitrary nonnegative real numbers satisfying r1 + r2 � 1.

By Lemma 7, we need to prove the follow inequality:

F 0r = 4(r1 + r2 + r3)4(r1t

51 + r2t

52) − (r1 + r2 + r3)3(r1t

41 + r2t

42)(r1t1 + r2t2)

− 4(r1 + r2 + r3)3(r1t31 + r2t

32)(r1t

21 + r2t

22) + 2(r1 + r2 + r3)2(r1t

31 + r2t

32)(r1t1 + r2t2)2

− 3(r1 + r2 + r3)2(r1t21 + r2t

22)

2(r1t1 + r2t2) + 3(r1 + r2 + r3)(r1t21 + r2t

22)(r1t1 + r2t2)3

− (r1t1 + r2t2)5 � 0,

where ∀t = (t1, t2) ∈ R2+, r1, r2, r3 are arbitrary nonnegative real numbers.

Call successive difference substitution program “tsds”. It returns “true” in less than 2 s. Thus for allpositive integers n, and ∀xi ∈ R+(i = 1, . . . , n), f0 � 0 holds. Therefore, G � 0 holds.

By Lemma 8, compute stationary point equations of f0

df0

dxi= 20n4x4

i − 4n3A1x3i − (12n3A2 − 6n2A2

1)x2i − (8n3A3 + 12n2A1A2 − 6nA3

1)xi

− (n3A4 − 4n2A1A3 + 3n2A22 − 9nA2

1A2 + 5A41) = 0, i = 1, . . . , n.

The degree of the above stationary point equations at most is 4. So every equation has no more than4 positive real roots. By Lemma 8, we need to prove the following inequality:

F 0r = 4

( 4∑i=1

ri

)4 4∑i=1

rit5i −

( 4∑i=1

ri

)3 4∑i=1

rit4i

4∑i=1

riti − 4( 4∑

i=1

ri

)3 4∑i=1

rit3i

4∑i=1

rit2i

+ 2( 4∑

i=1

ri

)2 4∑i=1

rit3i

( 4∑i=1

riti

)2

− 34∑

i=1

ri

( 4∑i=1

rit2i

)2 4∑i=1

riti + 34∑

i=1

ri

4∑i=1

rit2i

( 4∑i=1

riti

)3

−( 4∑

i=1

riti

)5

� 0,

where ∀t = (t1, . . . , t4) ∈ R4+, ri, i = 1, . . . , 4, are arbitrary nonnegative real numbers.

If we call successive difference substitution program “tsds”, it returns “true” after 6055 s. If we callpartial difference substitution program, and enter command “sds(F 0

r , {x1, x2, x3, x4})”, it returns “true”in less than 2 s. Thus for any positive integer n, and ∀xi ∈ R+(i = 1, . . . , n), f0 � 0 holds.

Remark 5. In this example, though sign variations of the coefficients sequence is hard to decide, westill can determine the upper bound of the number of positive real roots according the degree of theequation.

Remark 6. The decision of F 0r � 0 in this example is also considered in [15] using successive difference

substitution program.

Example 3. Assume that g(s) is integrable on interval [0, 1], and consider the following integral in-equality:

G =∫ 1

0

g8(s)ds −( ∫ 1

0

|g(s)|ds

)8

� 0.

Does G � 0 hold for all g(s) integrable on interval [0, 1]. Equivalently, consider the following polynomialinequality:

f0 =n∑

l=1

x8i

n−

( n∑i=1

xi

n

)8

� 0.

Does f0 � 0 hold for all positive integers n and all xi ∈ R+, i = 1, . . . , n?


By Lemma 6, we need to verify the following inequality:

f0r =

4∑i=1

rit8i −

( 4∑i=1

riti

)8

� 0,

where t = (t1, t2, r3, r4) ∈ R4+, r1, r2, r3, r4 are arbitrary nonnegative real numbers satisfying r1 + r2 +

r3 + r4 � 1.By Lemma 7, we need to verify the following inequality:

F 0r =

( 5∑i=1

ri

)7 4∑i=1

rit8i −

( 4∑i=1

riti

)8

� 0,

where ∀t = (t1, . . . , t4) ∈ R4+, r1, r2, r3, r4, r5 are arbitrary nonnegative real numbers.

Call partial difference substitution program. Enter command “sds(F 0r , {x1, x2, x3, x4})”. It returns

“true” after 57 s. Thus for any positive integer n, and ∀xi ∈ R+(i = 1, . . . , n), f0 � 0 holds.By Lemma 8, compute stationary point equations of f0:

df0

dxi= 8n7x7

i − 8A71 = 0, i = 1, 2, . . . , n.

According to Descartes’ law of signs, every stationary point equation has at most one nonnegative realroot. By Lemma 8, we only need to verify the following inequality:

F 0r = (r7

1)r1t81 − (r1t1)8 � 0,

where ∀t = (t1) ∈ R+, r1 is a nonnegative real number. Obviously, F 0r � 0. Thus for all positive integers

n, and ∀xi ∈ R+(i = 1, · · · , n), f0 � 0 holds. Therefore G � 0 holds.

Example 4. Assume that g(s) is integrable on interval [0, 1], and consider the following integral in-equality:

G = − 20(∫ 1

0

g4(s)ds

)2

+ 3∫ 1

0

|g(s)|3ds

∫ 1

0

|g(s)|5ds

+ 13∫ 1

0

g2(s)ds

∫ 1

0

g6(s)ds +∫ 1

0

g8(s)ds + 7∫ 1

0

|g(s)|7ds

∫ 1

0

|g(s)|ds � 0.

Does G � 0 hold for all g(s) integrable on interval [0, 1]? Equivalently, consider the following polynomialinequality:

f0 = −20( n∑

i=1

x4i

n

)2

+ 3n∑

i=1

x3i

n

n∑i=1

x5i

n+ 13

n∑i=1

x2i

n

n∑i=1

x6i

n+

n∑i=1

x8i

n+ 7

n∑i=1

x7i

n

n∑i=1

xi

n� 0.

Does f0 � 0 hold for all positive integers n and all xi ∈ R+, i = 1, · · · , n?By Lemma 6, we need to verify the following inequality:

f0r = −20

( 4∑i=1

rix4i

)2

+ 34∑

i=1

rix3i

4∑i=1

rix5i + 13

4∑i=1

rix2i

4∑i=1

rix6i +

4∑i=1

rix8i + 7

4∑i=1

rix7i

4∑i=1

rixi � 0,

where t = (t1, t2, r3, r4) ∈ R4+, r1, r2, r3, r4 are arbitrary nonnegative real numbers satisfying r1 + r2 +

r3 + r4 � 1.

By Lemma 7, we need to verify the following inequality:

F 0r = − 20

( 5∑i=1

ri

)6( 4∑i=1

rix4i

)2

+ 3( 5∑

i=1

ri

)6 4∑i=1

rix3i

5∑i=1

rix5i + 13

( 5∑i=1

ri

)6 4∑i=1

rix2i

4∑i=1

rix6i


+( 5∑

i=1

ri

)7 4∑i=1

rix8i + 7

( 5∑i=1

ri

)6 4∑i=1

rix7i

4∑i=1

rixi � 0,

where ∀t = (t1, . . . , t4) ∈ R4+, r1, r2, r3, r4, r5 are arbitrary real numbers.

After removing the positive common factor (∑5

i=1 ri)6 of F 0r , call partial difference substitution pro-

gram “sds”, and enter command “sds(F 0r , {x1, x2, x3, x4})”. It returns “true” in less than 1 s. Thus for

all positive integers n, and ∀xi ∈ R+(i = 1, . . . , n), f0 � 0 holds.By Lemma 8, compute the stationary point equations of f0:

df0

dxi= 8nx7

i + 49A1x6i + 78A2x

5i + 15A3x

4i − 160A4x

3i + 9A5x

2i + 26A6xi + 7A7 = 0, i = 1, . . . , n.

According to Descartes’ law of signs, every stationary point equation has no more than 2 nonnegativereal roots. Thus we need to prove the following inequality:

F 0r = − 20(r1 + r2)6(r1t

41 + r2t

42)

2 + 3(r1 + r2)6(r1t31 + r2t

32)(r1t

51 + r2t

52)

+ 13(r1 + r2)6(r1t21 + r2t

22)(r1t

61 + r2t

62) + (r1 + r2)7(r1t

81 + r2t

82)

+ 7(r1 + r2)6(r1t1 + r2t2)(r1t71 + r2t

72) � 0,

where ∀t = (t1, t2) ∈ R2+, r1, r2 are arbitrary nonnegative real numbers.

Call successive difference substitution program “tsds”. It returns “true” in less than 1 s. Thus for allpositive integers n, and ∀xi ∈ R+(i = 1, . . . , n), f0 � 0 holds. Therefore G � 0 holds.

Example 5. On the Website of Chinese Society of Inequalities and Applications, Chen claimed thatthe inequality proposed by Shi S1S2S7+S4S

23 � 2nS4S6 holds inversely, where Sk = xk

1+xk2+· · ·+xk

n, xi >

0, k = 1, 2, · · · , 7. We try to verify this claim by using our algorithm, but we have a counterexample:

n = 902, x1 = x2 = · · · = x899 = 1, x900 = x901 = x902 = 2.

This means the inequality proposed by Shi does not hold inversely. Later on, Chen verified that theinequality S1S2S7 + S4S

23 � 2nS4S6 holds when n < 1061). This example also means that the following

integral inequality

G =∫ 1

0

|g(s)|ds

∫ 1

0

g2(s)ds

∫ 1

0

|g(s)|7ds +∫ 1

0

g4(s)ds

( ∫ 1

0

|g(s)|3ds

)2

− 2∫ 1

0

g4(s)ds

∫ 1

0

g6(s)ds � 0

does not hold for all g(s) that are integrable on interval [0, 1].

Remark 7. Example 5 shows that when the given homogeneous integral inequality does not hold, ouralgorithm can give a numerical counterexample. It should be pointed out that the capability of givingcounterexample is due to successive difference substitution program.

5 Conclusions

In this paper, the problem of how to determine the nonnegativity for a class of integral inequalities isconsidered. The integral inequality is transformed into homogeneous symmetric polynomial inequalitybeyond Tarski model. The corresponding polynomial can be written as a new polynomial whose everyelement is an average power sum of the original variables. Using Timofte’s dimension-decreasing method,combined with determining polynomial non-negativeness method by use of the inequality proving softwarepackage BOTTEMA or a program which implements the method known as successive difference substi-tution, the nonnegativity of the corresponding polynomial inequality can be decided, hence the original

1) See http://old.irgoc.org/bbs/dispbbs.asp?boardid=17&id=2808&page=&star=2


integral inequality is mechanically decidable. A program “nm0prove” written in maple is developed andsome examples are provided to show the effectiveness of the algorithm.

Acknowledgements

This work was supported by the National Natural Science Foundation of China (Grant Nos. 60874010, 60572056,

90718041), the Knowledge Innovation Program of the Chinese Academy of Sciences (Grant No. KJCX-YW-

S02), and the Overseas Outstanding Young Researcher Foundation of the Chinese Academy of Sciences. The

authors also would like to thank Dr. Yang Zhengfeng and the anonymous referee for their helpful corrections and

suggestions which greatly improved the presentation of the paper.

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