ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY

ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY XIAO-MING ZHANG, BO-YAN XI, AND YU-MING CHU Abstract. This pap er gives another new proof method of analytic inequality involving n variables. As its applications, we give proof of some known-well inequalities and prove five new analytic inequalities. 1. monotonicity on special variables Throughout the paper R denotes the set of real numbers and R + denotes the set of strictly positive real numbers. Let n ≥ 2, n ∈ N. The arithmetic mean A(x) and the power mean M r (x) of order r with respect to the positive real numbers x 1 , x 2 , ··· , x n are defined respectively as A(x) = 1 n  n i=1 x i , M r (x) =  1 n  n i=1 x r i  1/r for r = 0, and M 0 (x) = (  n i=1 x i ) 1/n . In paper [4], the author puts up a new proof method of analytic inequality. In this section, we shall provide another new proof method of analytic inequality involving n variables. Lemma 1.1. Let interval I = [m, M] ⊂ R, D def. = {(x 1 , x 2 ) |m ≤ x 2 ≤ x 1 ≤ M} ⊂ I 2 and function f : I 2 → R have continuous partial derivative. Then ∂f/∂x 1 ≥ (≤) ∂f /∂x 2 hold in D, if and only if f (a, b) ≥ (≤) f (a − l, b + l ) hold for all a, b ∈ I and l with b < b + l ≤ a − l < a. Proof. Without the losing of generality, we only prove the c ase ∂f/∂x 1 ≥ ∂f/∂x 2 . For all x 1 , x 2 ∈ D and l ∈ R + with m ≤ x 2 < x 2 +l ≤ x 1 −l < x 1 ≤ M, we have f (x 1 − l, x 2 + l)− f (x 1 , x 2 ) ≤ 0. Then it exists ξ l ∈ (0, l) such that l  − ∂f (x 1 − ξ l , x 2 + ξ l ) ∂x 1 + ∂f (x 1 − ξ l , x 2 + ξ l ) ∂x 2  ≤ 0, − ∂f (x 1 − ξ l , x 2 + ξ l ) ∂x 1 + ∂f (x 1 − ξ l , x 2 + ξ l ) ∂x 2 ≤ 0. Let l → 0+, we get ∂f (x 1 , x 2 ) ∂x 1 ≥ ∂f (x 1 , x 2 ) ∂x 2 . According to continuity of partial derivative, we know ∂f (x 1 , x 1 ) ∂x 1 ≥ ∂f (x 1 , x 1 ) ∂x 2 . hold also. Date: September 13, 2009. 2000 Mathematics Subject Classification. Primary 26A48, 26B35, 26D20, Key words and phrases. monotone, maximum, minimum, inequality. This paper was typeset using A M S-L A T E X. 1 2 X M. ZHANG, B Y. XI, AND Y M. CHU On the other hand, assumes ∂f/∂x 1 ≥ ∂f/∂x 2 hold in D. For all a, b ∈ I and l with b < b + l ≤ a − l < a, it exists ξ l ∈ (0, l) such that f (a, b) −f (a −l, b + l) = −(f (a −l, b + l) − f (a, b)) = −l  − ∂f (a − ξ l , b + ξ l ) ∂x 1 + ∂f (a − ξ l , b + ξ l ) ∂x 2  = l  ∂f (a − ξ l , b + ξ l ) ∂x 1 − ∂f (a − ξ l , b + ξ l ) ∂x 2  ≥ 0. we complete the proof of lemma.  Theorem 1.1. (Compressed independent variables theorem) Suppose that D ⊂ R n be a symmetric with respect to permutations and convex set, and it have a nonempty interior set D 0 , function f : D → R and its partial derivatives be continue, and (1.1)  D i =  x ∈ D|x i = max 1≤j≤n {x j }  − {x ∈ D|x 1 = x 2 = ··· = x n }, (1.2)  D i =  x ∈ D|x i = min 1≤j≤n {x j }  − {x ∈ D|x 1 = x 2 = ··· = x n }, i = 1, 2, ··· , n. For all i, j = 1, 2, ··· , n, i = j, ∂f /∂x i > (<) ∂f/∂x j in  D i ∩  D j . Then (1.3) f (a 1 , a 2 , ··· , a n ) ≥ (≤) f (A (a) , A (a) , ··· , A (a)) for all a = (a 1 , a 2 , ··· , a n ) ∈ D, equality hold if only if a 1 = a 2 = ··· = a n . Proof. If n = 2, Let l → |a 1 − a 2 |/2 in Lemma 1.1, we complete the proof of theorem. We suppose n ≥ 3. Without the losing of generality, we only prove the case ∂f/∂x i > ∂f/∂x j with i = j. If a 1 = a 2 = ··· = a n , the inequality (1.3) hold obviously. If max 1≤j≤n {a j } = min 1≤j≤n {a j }, we let a 1 = max 1≤j≤n {a j } and a n = min 1≤j≤n {a j }. (1) If a 1 > max 2≤j≤n {a j }, a n < min 1≤j≤n−1 {a j }, because ∂f/∂x i > ∂f/∂x j hold in  D 1 ∩  D n , According to Lemma 1.1, exist a (1) 1 , a (1) n such that l = a 1 −a (1) 1 = a (1) n −a n > 0 and a (1) 1 = a i 0 = max 2≤j≤n−1 {a j } (let a (1) 1 = a 2 briefly ), or a (1) n = a j 0 = min 2≤j≤n−1 {a j } (let a (1) n = a n−1 briefly ), we have f (a 1 , a 2 , a 3 , ··· , a n ) ≥ f  a (1) 1 , a 2 , a 3 , ··· , a (1) n  . Simply, we denote a (1) i = a i , 2 ≤ i ≤ n − 1. Consequently, f (a 1 , a 2 , a 3 , ··· , a n ) ≥ f  a (1) 1 , a (1) 2 , a (1) 3 , ··· , a (1) n  . If a (1) 1 = a (1) 2 = ··· = a (1) n , implies Theorem 1.1 hold. Otherwise, for a (1) 1 = a (1) 2 > a (1) n , owing to ∂f (x) ∂x 1     x=  a (1) 1 ,a (1) 2 ,a (1) 3 ,··· ,a (1) n  > ∂f (x) ∂x n     x=  a (1) 1 ,a (1) 2 ,a (1) 3 ,··· ,a (1) n  , ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY 3 and the continuity of partial derivatives, it exists ε > 0 such that ∂f (x) ∂x 1     x=  s,a (1) 2 ,a (1) 3 ,··· ,t  > ∂f (x) ∂x n     x=  s,a (1) 2 ,a (1) 3 ,··· ,t  , where s ∈  a (1) 1 − ε, a (1) 1  t ∈  a (1) n , a (1) n + ε  . Denote a (2) 1 = a (1) 1 − ε, a (2) n = a (1) n + ε, a (2) i = a (1) i (2 ≤ i ≤ n − 1). By Lemma 1.1, we get (1.4) f  a (1) 1 , a (1) 2 , a (1) 3 , ··· , a (1) n  ≥ f  a (2) 1 , a (2) 2 , a (2) 3 , ··· , a (2) n  , and a (2) 2 = max 1≤i≤n  a (2) i  . For a (1) 1 > a (1) n−1 = a (1) n , after a similar argument, we get inequality (1.4) and a (2) n−1 = min 1≤i≤n  a (2) i  . Repeated the above steps, we get  a (i) 1 , a (i) 2 , ··· , a (i) n  (i = 1, 2, ···),  n j=1 a (i) j are constant,  a (i) j  (i = 1, 2, ···) are monotone increasing (decreasing) sequences if a j ≥ (≤) A (a) , j = 1, 2, 3, ··· , n, and f  a (1) 1 , a (1) 2 , a (1) 3 , ··· , a (1) n  ≥ f  a (i) 1 , a (i) 2 , a (i) 3 , ··· , a (i) n  . If exists i ∈ N, a (i) 1 = a (i) 2 = ··· = a (i) n , we complete the proof of theorem. Otherwise, let α = inf i∈N  max  a (i) 1 , a (i) 2 , ··· , a (i) n  . Without the losing of generality, we suppose max  a (i j ) 1 , a (i j ) 2 , ··· , a (i j ) n  = a (i j ) 1 → α and lim j→+∞  a (i j ) 1 , a (i j ) 2 , ··· , a (i j ) n  = (α, b 2 , b 3 , ··· , b n ) , where {i j } +∞ j=1 is a subsequence of N. Because of the continuity of function f , it have f (a 1 , a 2 , a 3 , ··· , a n ) ≥ f (α, b 2 , b 3 , ··· , b n ) . If α = min {b 2 , b 3 , ··· , b n }, we can repeat the above arguments, this contradicts with the definition of α. Then α = b 2 = b 3 = ··· = b n .Owing to α +  n i=2 b i =  n i=1 a i , we have α = b 2 = b 3 = ··· = b n = A(a), the proof of Theorem 1.1 is completed. (2) For the case a 1 = max 2≤j≤n {a j }, or a n = min 1≤j≤n−1 {a j }, it have been proved in (1).  In particular, according to Theorem 1.1 the following corollary hold. Corollary 1.1. Suppose that D ⊂ R n is a symmetric with respect to permutations and convex set, and it has a nonempty interior set D 0 , function f : D → R is symmetric, f and its partial derivatives is continue. Let  D 1 =  x ∈ D|x 1 = max 1≤j≤n {x j }  − {x ∈ D|x 1 = x 2 = ··· = x n },  D 2 =  x ∈ D|x 2 = min 1≤j≤n {x j }  − {x ∈ D|x 1 = x 2 = ··· = x n }, (1.5) D ∗ =  D 1 ∩  D 2 . If ∂f/∂x 1 > (<) ∂f/∂x 2 hold in D ∗ , then f (a 1 , a 2 , ··· , a n ) ≥ (≤) f (A (a) , A (a) , ··· , A (xa)) for all a = (a 1 , a 2 , ··· , a n ) ∈ D, equality hold if only if a 1 = a 2 = ··· = a n . 4 X M. ZHANG, B Y. XI, AND Y M. CHU 2. Unifying Proof of Some Well-known Inequality Take advantage of Theorem 1.1 and Corollary 1.1, we can prove some well-known inequality, for example, Power Mean Inequality, Holder-Inequality, Minkowski-Inequality. In this section, we only prove an example. Proposition 2.1. (Holder-Inequality) Let (x 1 , x 2 , ··· , x n ), (y 1 , y 2 , ··· , y n ) ∈ R n + , p, q > 1, and 1/p + 1/q = 1. Then   n k=1 x p k  1/p   n k=1 y q k  1/q ≥  n k=1 x k y k . Proof. Let (a 1 , a 2 , ··· , a n ) ∈ R n + and function f : b →   n k=1 a k  1/p   n k=1 a k b k  1/q −  n k=1 a k b 1/q k , b ∈ R n + . We have ∂f ∂b i = 1/q ·   n k=1 a k  1/p   n k=1 a k b k  1/q−1 a i − 1/q · a i b 1/q−1 i , ∂f ∂b i − ∂f ∂b j = 1 q   n k=1 a k  n k=1 b k a k  1/p (a i − a j ) − 1 q  a i b −1/p i − a j b −1/p j  . Let b ∈  D i ∩  D j (see (1.1) and (1.2)). (1)If a i ≥ a j , ∂f ∂b i − ∂f ∂b j ≥ 1 q   n k=1 a k b i  n k=1 a k  1/p (a i − a j ) − 1 q  a i b −1/p i − a j b −1/p j  = 1 q a j  b −1/p j − b −1/p i  > 0. (2)If a i ≤ a j , ∂f ∂b i − ∂f ∂b j ≥ 1 q   n k=1 a k b j  n k=1 a k  1/p (a i − a j ) − 1 q  a i b −1/p i − a j b −1/p j  = 1 q a i  b −1/p j − b −1/p i  > 0. According to 1.1, we get f (b) ≥ f (A (b) , A (b) , ··· , A (b)) . that is (2.1)   n k=1 a k  1/p   n k=1 a k b k  1/q −  n k=1 a k b 1/q k ≥ 0. Let a k = x p k , , b k = y q k  x p k in (2.1), we know Holder-Inequality hold.  3. Five new inequalities Let n ≥ 3, a = (a 1 , a 2 , ··· , a n ) ∈ R n + ,  k n (a) =   1≤i 1 <···<i k ≤n 1 k  k j=1 a i j  1  n k  is the third symmetric mean of a (see [2]). Theorem 3.1. Let 2 ≤ k ≤ n − 1, p = (k −1)/(n − 1). Then (3.1)  k n (a) ≥ [A (a)] p [M 0 (a)] 1−p , p = (k −1)/(n − 1) is the best constant. ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY 5 Proof. Suppose f (a) =   n i=1 a i  − (n−k)·    n k    n(n−1) ·  1≤i 1 <···<i k ≤n 1 k  k j=1 a i j , a = (a 1 , a 2 , ··· , a n ) ∈ R n + . Then ∂f ∂a 1 = − (n − k) ·  n k  n (n − 1) a 1   n i=1 a i  − (n−k)·    n k    n(n−1) ·  1≤i 1 <···<i k ≤n 1 k  k j=1 a i j , ∂f ∂a 1 − ∂f ∂a 2 = − (n − k) ·  n k  n (n − 1)   n i=1 a i  − (n−k)·    n k    n(n−1) ·  1≤i 1 <···<i k ≤n 1 k  k j=1 a i j  1 a 1 − 1 a 2  = (n − k) ·  n k  n (n − 1) a 1 a 2   n i=1 a i  − (n−k)·    n k    n(n−1) ·  1≤i 1 <···<i k ≤n 1 k  k j=1 a i j (a 1 − a 2 ) −   n i=1 a i  − (n−k)·    n k    n(n−1) ·  1≤i 1 <···<i k ≤n 1 k  k j=1 a i j ·  3≤i 1 <···i k−1 ≤n a 1 − a 2  a 1 +  k−1 j=1 a i j  a 2 +  k−1 j=1 a i j  . (3.2) If a ∈ D ∗ (see (1.5)), a 1 + (k − 1) a 2 > a 1 , (n − k) ·  n k  n (n − 1) a 1 a 2 >  n − 2 k − 1  ka 2 (a 1 + (k − 1) a 2 ) , (n − k) ·  n k  n (n − 1) a 1 a 2 >  3≤i 1 <···i k−1 ≤n 1 (a 1 + (k − 1) a 2 ) ka 2 , (3.3) (n − k) ·  n k  n (n − 1) a 1 a 2 >  3≤i 1 <···i k−1 ≤n 1  a 1 +  k−1 j=1 a i j  a 2 +  k−1 j=1 a i j  . Combining inequality (3.2) to inequality (3.3), we have ∂f/∂a 1 −∂f/∂a 2 > 0. Owing to Corollary 1.1, we get f (a 1 , a 2 , ··· , a n ) ≥ (≤) f (A (a) , A (a) , ··· , A (a)) 6 X M. ZHANG, B Y. XI, AND Y M. CHU for all a = (a 1 , a 2 , ··· , a n ) ∈ R n + . It implies   n i=1 a i  − (n−k)·    n k    n(n−1) ·  1≤i 1 <···<i k ≤n k −1  k j=1 a i j ≥ [A (a)] (k−1)·    n k    n−1 . Inequality (3.1) is proved. Let a 1 = a 2 = ··· = a n−1 = 1, a n = x in inequality (3.1), it lead to  x + k − 1 k   n − 1 k − 1  n k  ≥  x + n − 1 n  p  n √ x  1−p . p ≤ k n ln x+k−1 k − 1 n ln x ln (x + n − 1) − ln n n √ x . Let x → +∞ in above inequality, p ≤ lim x→+∞ k n · 1 x+k−1 − 1 nx 1 x+n−1 − 1 nx = lim x→+∞ kx x+k−1 − 1 nx x+n−1 − 1 = k − 1 n − 1 . So p = (k − 1)/(n − 1) is the best constant.  Theorem 3.2. Suppose n ≥ 3, a = (a 1 , a 2 , ··· , a n ) ∈ R n + , β > 0 > α. If β +α > 0, let λ = −2α n(β−α) . If β + α ≤ 0, let λ = 1 n . Then (3.4) [M α (a)] 1−λ · [M β (a)] λ ≤ M 0 (a) . Proof. Let f (x) = 1 nβ ln (  n i=1 x i ) − 1−λ α ln  1 n  n i=1 x α/β i  , x ∈ R n + . Then ∂f (x) ∂x j = 1 nβx j − 1 − λ β x α/β−1 j  n i=1 x α/β i , j = 1, 2, (3.5) ∂f (x) ∂x 1 − ∂f (x) ∂x 2 = x 2 − x 1 nβx 1 x 2 − 1 − λ β x α/β−1 1 − x α/β−1 2  n i=1 x α/β i . Case 1 :α + β > 0. Let g (t) = β + α β −α t β−α − t β + t −α − β + α β −α , t ∈ (1, +∞) . Then t α+1 g  (t) = (β + α) t β − βt β+α − α,  t α+1 g  (t)   = (β + α) βt β+α−1  t −α − 1  > 0. Therefore t α+1 g  (t) is monotone increasing function in (1, +∞). Meanwhile lim t→1+ t α+1 g  (t) = lim t→1+  (β + α) t β − βt β+α − α  = 0. Thence, t α+1 g  (t) > 0, g  (t) > 0. In addition to lim t→1+ g (t) = 0, we have g (t) > 0 and β + α β −α t β−α − t β + t −α − β + α β −α > 0 β + α β −α t β −  1 + 2α β −α  t α − t α+β + 1 > 0, ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY 7 β + α β −α t β −  n − 1 + 2α β −α  t α − t α+β + (n − 1) > 0, (3.6) (1 − nλ) t β − (n − 1 − nλ) t α − t α+β + (n − 1) > 0, (3.7) (1 − λ) 1 − t α−β t α + (n − 1) > t β − 1 nt β . We assume that x ∈ D ∗ (see (1.5)), let t = (x 1 /x 2 ) 1/β in above inequality. It has (1 − λ) x α/β−1 2 − x α/β−1 1 x α/β 1 + (n − 1) x α/β 2 > x 1 − x 2 nx 1 x 2 , (3.8) 1 − λ β x α/β−1 2 − x α/β−1 1  n i=1 x α/β i > x 1 − x 2 nβx 1 x 2 . Combining inequality (3.5) to inequality (3.8), we have ∂f (x)/∂x 1 − ∂f (v)/∂x 2 > 0. According to Corollary 1.1, we get f (x 1 , x 2 , ··· , x n ) ≥ f (A (x) , A (x) , ··· , A (x)) , 1 nβ ln   n i=1 x i  − 1 − λ α ln  1 n  n i=1 x α/β i  ≥ λ β ln  1 n  n i=1 x i  . Let a i = x 1/β i , i = 1, 2, ··· , n, it get [M α (a)] 1−λ · [M β (a)] λ ≤ M 0 (a) . Case 2: α + β < 0. Let t > 1, because α < 0, α + β < 0, it has (n − 1) > (n − 2) t α + t α+β . We find that inequality (3.7) hold.The rest of the pro ofs are similar, so we shall omit them.  A similar argument lead to Theorem 3.3 Theorem 3.3. Suppose n ≥ 3, a = (a 1 , a 2 , ··· , a n ) ∈ R n + , β > 0 > α. If β + α > 0, let θ = n−1 n . If β + α ≤ 0, let θ = 1 − 2β n(β−α) . Then (3.9) M 0 (a) ≤ [M α (a)] 1−θ · [M β (a)] θ . Remark 3.1. Inequality 3.4 3.9 improve the well-known Sierpinski-Inequality [M −1 (a)] n−1 n [A (a)] 1 n ≤ M 0 (a) ≤ [M −1 (a)] 1 n [A (a)] n−1 n . For n ≥ 2, a = (a 1 , a 2 , ··· , a n ) ∈ R n + , paper [1] introduces an inequality n − 1 n A (a) + 1 n M −1 (a) ≥ M 0 (a) We can improve the inequality by Theorem 3.4 and Theorem 3.5. Theorem 3.4. Suppose p = n 2  n 2 + 4n − 4  , then (3.10) pA (a) + (1 − p) M −1 (a) ≥ M 0 (a) . 8 X M. ZHANG, B Y. XI, AND Y M. CHU Proof. Firstly, Let p > n 2  n 2 + 4n − 4  , f (b) = p/n ·  n i=1 e b i + (1 − p) n ·   n i=1 e −b i  −1 , b = (b 1 , b 2 , ··· , b n ) ∈ R n . Then ∂f ∂b 1 = p n e b 1 + (1 − p) n (  n i=1 e −b i ) 2 e −b 1 , ∂f ∂b 1 − ∂f ∂b 2 = p n  e b 1 − e b 2  − (1 − p) n (  n i=1 e −b i ) 2  e −b 2 − e −b 1  . If b 1 = max 1≤i≤n {b i } > b 2 = min 1≤i≤n {b i }, t = e b 1 −b 2 > 1. We have ∂f ∂b 1 − ∂f ∂b 2 ≥ p n  e b 1 − e b 2  − (1 − p) n ((n − 1) e −b 1 + e −b 2 ) 2  e −b 2 − e −b 1  =  e b 1 − e b 2  n ((n − 1) e b 2 + e b 1 ) 2  p  (n − 1) e b 2 + e b 1  2 − (1 − p) n 2 e b 1 e b 2  = e 3b 2 (t − 1) n ((n − 1) e b 2 + e b 1 ) 2  p (n − 1 + t) 2 − n 2 t + pn 2 t  . > e 3b 2 (t − 1) n ((n − 1) e b 2 + e b 1 ) 2  n 2 n 2 + 4n − 4 (n − 1 + t) 2 − n 2 t + n 2 n 2 + 4n − 4 n 2 t  = ne 3b 2 (t − 1) (t − n + 1) 2 (n 2 + 4n − 4) ((n − 1) e b 2 + e b 1 ) 2 ≥ 0. According to Corollary 1.1, we get f (b) ≥ f (A (b) , A (b) , ··· , A (b)) , p n  n i=1 e b i + (1 − p) n  n i=1 e −b i ≥ e A(b) = n   n i=1 e b i . Let e b i = a i in above inequality, we know inequality (3.10) hold. Because of continuity, if p = n 2  n 2 + 4n − 4  , inequality (3.10) hold also .  Theorem 3.5. Suppose p =  1 − n − √ 5n 2 − 6n + 1  2n, then n − 1 n A (a) + 1 n M p (a) ≥ M 0 (a) . Proof. Let f (a) = n   n i=1 a 1/p i − (n − 1)  n 2 ·  n i=1 a 1/p i , a ∈ R n + . Then ∂f ∂a 1 = 1 npa 1 n   n i=1 a 1/p i − n − 1 n 2 p a 1/p−1 1 , ∂f ∂a 1 − ∂f ∂a 2 = − a 1 − a 2 npa 1 a 2  n i=1 a 1/np i − n − 1 n 2 p  a 1/p−1 1 − a 1/p−1 2  . If a 1 = max 1≤i≤n {a i } > a 2 = min 1≤i≤n {a i } > 0, a 1 /a 2 = t > 1. Owing to p < 0,− a 1 −a 2 npa 1 a 2 > 0, we have ∂f ∂a 1 − ∂f ∂a 2 ≤ − a 1 − a 2 npa 1 a 2 a 1/(np) 1 a (n−1)/(np) 2 − n − 1 n 2 p  a 1/p−1 1 − a 1/p−1 2  = a 1/p−1 1 n 2 p  −n t − 1 t t 1−(n−1)/(np) − (n − 1)  1 − t 1−1/p   . (3.11) ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY 9 Let g (t) = −nt 1−(n−1)/(np) + nt −(n−1)/(np) + (n − 1) t 1−1/p − (n − 1) , t > 1. g  (t) =  −n + n − 1 p  t −(n−1)/(np) − n − 1 p t −1−(n−1)/(np) + (n − 1)  1 − 1 p  t −1/p , t 1+(n−1)/(np) g  (t) =  −n + n − 1 p  t − n − 1 p + (n − 1)  1 − 1 p  t 1−1/(np) .  t 1+(n−1)/(np) g  (t)   =  −n + n − 1 p  + (n − 1)  1 − 1 p  1 − 1 np  t −1/(np) >  −n + n − 1 p  + (n − 1)  1 − 1 p  1 − 1 np  = 0. Thus t 1+(n−1)/(np) g  (t) is a monotone increasing function. Meanwhile, lim t→1+ t 1+(n−1)/(np) g  (t) = lim t→1+  −n + n − 1 p  t − n − 1 p + (n − 1)  1 − 1 p  t 1− 1 np  = −1 − n − 1 p ≥ 0. Therefore t 1+(n−1)/(np) g  (t) > 0, g  (t) > 0, g (t) is monotone increasing function. Meanwhile, lim t→1+ g (t) = 0, then g (t) > 0. By (3.11), we know ∂f/∂a 1 − ∂f/∂a 2 < 0. According to Corollary 1.1, f (a) ≤ f (A (a) , A (a) , ··· , A (a)) , n   n i=1 a 1/p i − n − 1 n 2 ·  n i=1 a 1/p i ≤ 1 n · A 1/p (a) Finally, let a i → a p i (i = 1, 2, ··· , n) in the above inequality, we know Theorem 3.5 hold.  Remark 3.2. More applications of Theorem 1.1 and Corollary 1.1 will appear in book [5] and http://old.irgoc.org/Article/ShowArticle.asp?ArticleID=391. Acknowledgments This work was supported by the NSF of China Central Radio and TV University under Grant No. GEQ1633 and Foundation of the Educational Committee of Zhejiang Province under Grant Y200804124. References [1] Alzer, H., Sierpinski’s inequality, J. Belgian Math. Soc. B, 41 (1989), 139-144. [2] J J. Wen, Hardy mean and their inequalities, J. of Math., 27, 4 (2007), 447–450(in chinese). [3] J J. Wen, W L. Wang, The optimizations for the inequalities of power means. Joural of In- equalities and Applications, 2006 (2006), Article ID 46782, 25 page. [ONLINE]Available online at http://www.hindawi.com/journals/jia/volume-2006/regular.91.html [4] X M. Zhang, A new proof method of analytic inequality. Research Report Collection, 12, 1(2009), Art. 2. [ON- LINE] Available online at http://www.staff.vu.edu.au/RGMIA/v12n1.asp [5] X M. Zhang, Y M. Chu, New discussion to analy tic inequality, HarBin: HarBin Institute of Technology Press, 2009. (in chinese) [6] X M. Zhang, Optimization of Schur-Convex Functions, Mathematical Inequalities and Applications, 1, 3 (1998), 319-330. [ONLINE] Available online at http://www.mia-journal.com/miasup.asp 10 X M. ZHANG, B Y. XI, AND Y M. CHU (X M. Zhang) Zhejiang Broadcast and TV University Haining College, Haining City, Zhejiang Province, 314400, P. R. China E-mail address: zjzxm79@126.com (B Y. Xi) Department of Mathematics of Inner Mongolia University for the Nationalities, Tongliao, Inner Mongolia, 028000, P. R. China E-mail address: baoyintu68@sohu.com (Y M. Chu) Department of Mathematics, Huzhou Teachers C ollege, Huzhou 313000, P.R.China E-mail address: chuyuming@hutc.zj.cn . ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY XIAO-MING ZHANG, BO-YAN XI, AND YU-MING CHU Abstract. This pap er gives another new proof method of analytic inequality. (  n i=1 x i ) 1/n . In paper [4], the author puts up a new proof method of analytic inequality. In this section, we shall provide another new proof method of analytic inequality involving n variables. Lemma. Then (3.1)  k n (a) ≥ [A (a)] p [M 0 (a)] 1−p , p = (k −1)/(n − 1) is the best constant. ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY 5 Proof. Suppose f (a) =   n i=1 a i  − (n−k)·    n k    n(n−1) ·  1≤i 1 <···<i k ≤n 1 k  k j=1 a i j ,