A reconsideration of Hua's inequality
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The inequality discovered by L. K. Hua in 1965 has been generalized in several directions. In this paper, we adopt a certain conjugate method to give a simple and fundamental inequality on two functions on a semigroup, which is the key to the proof of many generalizations of Hua’s inequality. 1. Introduction In 1965, L. K. Hua discovered the following inequality which plays an important role in number theory. Theorem 1.1 (Hua’s inequality [2]). If δ,α > 0 and x1 ,...,xn ∈ R, then
δ−
n i=1
2
xi
+α
n i=1
xi2 ≥
αδ 2 . α+n
(1.1)
In (1.1), the equality holds if and only if x1 = · · · = xn = δ/(α + n). This inequality has been generalized as follows. Theorem 1.2 (Wang’s inequality [6]). If p > 1, then p n n p α p−1 p δ − xi + α p−1 xi ≥ δ α+n i=1
(1.2)
i=1
for any δ,α > 0, x1 ,...,xn ∈ R. If 0< p < 1, then the inequality sign in (1.2) is reversed, where δ,α > 0, x1 ,...,xn ≥ 0 with ni=1 xi ≤ δ. In both cases, the equality holds in (1.2) if and only if x1 = · · · = xn = δ/(α + n).
Copyright © 2005 Hindawi Publishing Corporation Journal of Inequalities and Applications 2005:1 (2005) 15–23 DOI: 10.1155/JIA.2005.15
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A reconsideration of Hua’s inequality
Theorem 1.3 (Dragomir-Yang inequalities [1]). Let X be a real inner product space. Then
2
δ − x, y + αx2 ≥
αδ 2 , α + y 2
(1.3)
2 n n
2 α
y − x i + α x i ≥ y 2
α + n i=1 i=1
(1.4)
for any δ,α > 0, x, y,x1 ,...,xn ∈ X. In (1.3), the equality holds if and only if x = (δ/(α + y 2 ))y. In (1.4), the equality holds if and only if x1 = · · · = xn = (1/(α + n))y. In [4], Pearce and Peˇcari´c pointed out that these inequalities are consequences of Jensen’s inequality. Moreover, Peˇcari´c [5] showed the following three inequalities. Theorem 1.4 (Peˇcari´c’s inequalities [5]). Let X be a real inner product space. If p, q > 1 and 1/ p + 1/q = 1, then δ − x, y p + α p−1 x p ≥
α
p−1
α + y q
δp
(1.5)
for any δ,α > 0, x, y ∈ X. The equality holds in (1.5) if and only if x = y = 0 or x = (δ y q−2 /(α + y q ))y. Moreover, if f is a nondecreasing convex function on [0, ∞), then
y α + y αδ f δ − x, y + f αx ≥ f , α α α + y
n n
1
xi ≥ α + n f α y f
y − xi + f α
α i=1 α α+n i=1
(1.6) (1.7)
for any δ,α > 0, x, y,x1 ,...,xn ∈ X. When f is strictly convex, the equality in (1.6) holds if and only if y = 0 or x = (δ/( y (α + y )))y, and the equality in (1.7) holds if and only if x1 = · · · = xn = (1/(α + n))y. In this paper, we use a certain conjugate method to give a more extended formulation including the above inequalities. 2. Results The main result of this paper is the following inequality. Theorem 2.1. Let (G,+) be a semigroup, and let ϕ and ψ be nonnegative functions on G. Suppose ϕ is subadditive on G and there is a positive constant λ such that ϕ(x) ≤ λψ(x) for x ∈ G. If f is a nondecreasing convex function on [0, ∞), then
f ϕ a0
+λ
n i=1
f ψ ai
≥ (1 + nλ) f
n 1
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