© Hindawi Publishing Corp. KY FAN INEQUALITY AND BOUNDS FOR DIFFERENCES OF MEANS

We prove an equivalent relation between Ky Fan-type inequalities 
and certain bounds for the differences of means. We also generalize a result of Alzer et al. (2001).


Introduction.
Let P n,r (x) be the generalized weighted power means: P n,r (x) = ( n i=1 ω i x r i ) 1/r , where ω i > 0, 1 ≤ i ≤ n with n i=1 ω i = 1 and x = (x 1 ,x 2 ,...,x n ).Here, P n,0 (x) = n i=1 x ω i i denotes the limit of P n,r (x) as r → 0 + , which can be proved by noting that if p(r ) = ln( n i=1 ω i x r i ), then p (0) = ln( n i=1 x ω i i ) = ln(P n,0 (x)).We write P n,r for P n,r (x) when there is no risk of confusion.
In this paper, we assume that 0 < x 1 ≤ x 2 ≤ ••• ≤ x n .With any given x, we associate x = (1 − x 1 , 1 − x 2 ,...,1 − x n ) and write A n = P n,1 , G n = P n,0 , and H n = P n,−1 .When 1− x i ≥ 0 for all i, we define A n = P n,1 (x ) and similarly for G n and H n .We also let σ n = n i=1 ω i [x i − A n ] 2 .The following counterpart of the arithmetic mean-geometric mean inequality, due to Ky Fan, was first published by Beckenbach and Bellman [7].
Theorem 1.1.For x i ∈ (0, 1/2], with equality holding if and only if In this paper, we consider the validity of the following additive Ky Fan-type inequalities (with x 1 < x n < 1): Note that by a change of variables x i → 1 − x i , the left-hand side inequality is equivalent to the right-hand side inequality in (1.2).We can deduce (see [9]) Theorem 1.1 from the case r = 1, s = 0, and x n ≤ 1/2 in (1.2), which is a result of Alzer [5].Gao [9] later proved the validity of (1.2) for r = 1, −1 ≤ s < 1, and What is worth mentioning is a nice result of Mercer [12] who showed that the validity of r = 1 and s = 0 in (1.2) is a consequence of a result of Cartwright and Field [8] who established the validity of r = 1 and s = 0 for the following bounds for the differences between power means (r > s): where the constant (r − s)/2 is the best possible (see [10]).We point out that inequalities (1.2) and (1.3) do not hold for all r > s.We refer the reader to the survey article [2] and the references therein for an account of Ky Fan's inequality, and to [4,5,10,11] for other interesting refinements and extensions of (1.3).
Mercer's result reveals a close relation between (1.3) and (1.2), and it is our main goal in the paper to prove that the validities of (1.3) and (1.2) are equivalent for fixed r and s.As a consequence of this result, we give a characterization of the validity of (1.3) for r = 1 or s = 1.A solution of an open problem from [11] is also given.
Among the numerous sharpenings of Ky Fan's inequality in the literature, we have the following inequalities connecting the three classical means (with ω i = 1/n here): The right-hand side inequality of (1.4) is due to W. L. Wang and P. F. Wang [14] and the left-hand side inequality was recently proved by Alzer et al. [6].
It is natural to ask whether we can extend the above inequality to the weighted case, and using the same idea as in [6], we show that this is indeed true in Section 5.

The main theorem
Theorem 2.1.For fixed r > s, the following inequalities are equivalent: Proof.(iii)⇒(ii) follows from a similar argument as given in [12], (ii)⇒(i) is trivial, so it suffices to show that (i)⇒(iii).
Fix r > s assuming that (1.2) holds for x n ≤ 1/2.Without loss of generality, we can assume that x 1 < x n .For a given x = (x 1 ,x 2 ,...,x n ), let y = ( x 1 , x 2 ,..., x n ).We can choose small so that x n ≤ 1/2.Now, applying the right-hand side inequality (1.2) for y, we get Let f ( ) = P n,r (y ) − P n,s (y ), then f (0) = 0 and f (0) = (r − s)σ n .Thus, by letting tend to 0, it is easy to verify that the limit of the expression on the right-hand side of (2.1) is (r − s)σ n /2.We can consider the left-hand side of (1.2) by a similar argument and this completes the proof.
We note here that a special case of Theorem 3.2 answers an open problem of Mercer [11], namely, we have shown that
Lemma 4.2.Let x, a, b, u, v, and s be real numbers with with equality holding if and only if one of the following cases is true: Furthermore, we define H(s, a) as the expression on the left-hand side of (4.3), where (s, a) ∈ M. It suffices to show that H(s, a) < 0. We denote the absolute minimum of H by m = (s 0 ,a 0 ).If m is an interior point of M, then we obtain Hence, m is a boundary point of M, so we get m ∈ s 0 ,x , s 0 ,b , 0,a 0 , v, a 0 .(4.5) Using Lemma 4.1, we obtain Thus, we get that if (s, a) ∈ M, then H(s, a) ≤ 0. The conditions for equality can be easily checked using Lemma 4.1.

A sharpening of Ky Fan's inequality.
In this section, we prove the following theorem.
with equality holding if and only if q = 1/2 or Proof.The proof uses the ideas in [6].We prove the right-hand side inequality of (5.1); the proofs for other inequalities are similar.Fix 0 < x = x 1 , x n = b with x 1 < x n , n ≥ 2; we define where we regard A n , G n , and H n as functions of x n = (x 1 ,...,x n ).
We then have We want to show that ..,a n−1 ) ∈ D be the point in which the absolute minimum of g n is reached.Next, we show that a = (x,...,x,a,...,a,b,...,b) with x < a < b, ( where the numbers x, a, and b appear r , s, and t times, respectively, with r ,s,t ≥ 0 and r + s + t = n − 2.

PENG GAO
Suppose not, this implies that two components of a have different values and are interior points of D. We denote these values by a k and a l .Partial differentiation leads to for i = k, l, where Since z B/z 2 + C is strictly monotonic for z > 0, then (5.6) yields a k = a l .This contradicts our assumption that a k ≠ a l .Thus, (5.5) is valid and it suffices to show that g n ≤ 0 for the case n = 2, 3.
We note that the above theorem gives a sharpening of Sierpiński's inequality [13], originally stated for the unweighted case (ω i = 1/n) as (5.10) The following corollary gives refinements of (1.4).

Corollary 5.2. For
with equality holding if and only if Proof.This is a direct consequence of Theorem 5.1, following from a similar argument as in [12].
It was conjectured that an additive companion of (1.4) is true (see [1]) In [3], Alzer asked if the above conjecture is true and whether there exists a weighted version.Based on what we have got in this paper, it is natural to give the following conjecture of the weighed version of (6.2).Conjecture 6.1.For 0 < x 1 ≤ ••• ≤ x n ≤ 1/2 and q = min{ω i },  (6.4) for all x i ∈ (0, 1/2] (i = 1,...,n)?
We note here that α ≤ 0 since the left-hand side inequality above can be written as (6.5)By a similar argument as in the proof of Theorem 2.1, replacing (x 1 ,...,x n ) by ( x 1 ,..., x n ) and letting tend to 0 in (6.5), we find that (6.5) implies that αA n + (1 − α)H n − G n ≤ 0 (6.6) for any x.If we further let x 1 tend to 0 in (6.6), we get αA n ≤ 0 (6.7) which implies that α ≤ 0.

. 3 )
Recently, Alzer et al.[6] asked the following question: what is the largest number α = α(n) and what is the smallest number β = β(n) such that