A reconsideration of Hua's inequality. Part II
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We give a new interpretation of Hua’s inequality and its generalization. From this interpretation, we know the best possibility of those inequalities. Copyright © 2006 Hiroyuki Takagi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction In 1965, L. Keng Hua discovered the following inequality. Theorem 1.1 [2]. If δ,λ > 0 and x1 ,...,xn ∈ R, then
δ−
n i =1
2
xi
+λ
n i=1
xi 2 ≥
λδ 2 . λ+n
(1.1)
In (1.1), the equality holds if and only if x1 = · · · = xn = δ/(λ + n). This inequality played an important role in number theory and has been generalized in several directions [1, 3–6]. One of its generalizations states the following. Theorem 1.2 [5, Corollary 2.7]. Let X be a real or complex normed space with dual X ∗ , and suppose p, q > 1 and 1/ p + 1/q = 1. If δ,λ > 0, x ∈ X, and f ∈ X ∗ , then δ − f (x) p + λ p−1 x p ≥
λ
λ + f q
p −1
δ p.
(1.2)
In (1.2), the equality holds if and only if f (x) = f x and x = δ f q−1 /(λ + f q ). In this paper, we give a new interpretation of the inequality (1.2) and consider whether the coefficients λ p−1 and (λ/(λ + f q )) p−1 are best possible. For this purpose, we divide
Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 21540, Pages 1–8 DOI 10.1155/JIA/2006/21540
2
A reconsideration of Hua’s inequality. Part II
both sides of (1.2) by (λ/(λ + f q )) p−1 δ p , and then replace x/δ by x. Thus we obtain a replica of Theorem 1.2. Theorem 1.3. Let X be a real or complex normed space with dual X ∗ , and suppose p, q > 1 and 1/ p + 1/q = 1. If λ > 0, x ∈ X, and f ∈ X ∗ , then
p −1
λ + f q λ
1 − f (x) p + λ + f q p−1 x p ≥ 1.
(1.3)
In (1.3), the equality holds if and only if f (x) = f x and x = f q−1 /(λ + f q ). Clearly, Theorems 1.2 and 1.3 are equivalent. So, we turn our attention to Theorem 1.3, which is more convenient for us. Put Ω=
1 − f (x), x : x ∈ X .
(1.4)
Then Ω is a subset of R+ × R+ , where R+ = {s ∈ R : s ≥ 0}. Moreover, we have
Ω ⊂ (s,t) ∈ R+ × R+ : s + f t ≥ 1 ,
(1.5)
because |1 − f (x)| + f x ≥ 1 − | f (x)| + f x ≥ 1 for all x ∈ X. While the inequality (1.3) has the form as p + bt p ≥ 1
∀(s,t) ∈ Ω,
(1.6)
where a and b are nonnegative constants. If we know all the nonnegative constants a and b such that (1.6) holds, then we may determine whether the coefficients ((λ + f q )/λ) p−1 and (λ + f q ) p−1 in (1.3) are best possible. 2. General theory Let k and be positive numbers. Let Ω be an index set such that
Ω ⊂ (s,t) ∈ R+ × R+ : ks + t ≥ 1 .
(2.1)
For such an index set Ω and any p > 0, we consider the domain
D(p;Ω) = (a,b) ∈ R+ × R+ : as p + bt p ≥ 1 ∀(s,t) ∈ Ω .
(2.2)
We wish to identify the domain D(p;Ω). We first consider the case p > 1. We define a function h p,k, on the open interval (
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