A NEW APPROACH TO KY FAN-TYPE INEQUALITIES

The study of the behavior of means under equal increments of their variables provides a new approach to Ky Fan-type inequalities. Via this approach we are able to prove some new results on Ky Fan-type inequalities. We also prove some inequalities involving the symmetric means.


Introduction
Let M n,r (x) be the generalized weighted power means: M 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 M n,0 (x) denotes the limit of M n,r (x) as r → 0 + .Unless specified otherwise, we always assume 0 < x 1 ≤ x 2 ••• ≤ x n .We denote σ n = n i=1 ω i (x i − A n ) 2 .To any given x, t ≥ 0 we associate x = (1 − x 1 ,1 − x 2 ,...,1 − x n ), x t = (x 1 + t,...,x n + t).When there is no risk of confusion, we will write M n,r for M n,r (x), M n,r,t for M n,r (x t ), and M n,r for M n,r (x ) if x n < 1.We also define A n = M n,1 , G n = M n,0 , H n = M n,−1 and similarly for A n , G n , H n , A n,t , G n,t , H n,t .
To simplify expressions, we define with ∆ r,s,t,0 = (ln(M n,r,t /M n,s,t ))/(ln(M n,r /M n,s )).We also write ∆ r,s,t for ∆ r,s,t,1 .In order to include the case of equality for various inequalities in our discussions, for any given inequality, we define 0/0 to be the number which makes the inequality an equality.Recently, the author [14, Theorem 2.1] has proved the following result.
Theorem 1.1.For r > s, the following inequalities are equivalent: where in (1.3) we require x n < 1.
Cartwright and Field [9] first proved the validity of (1.2) for r = 1, s = 0.For other extensions and refinements of (1.2), see [3,13,18,19].Inequality (1.3) is commonly referred to as the additive Ky Fan's inequality.We refer the reader to the survey article [2] and the references therein for an account of Ky Fan's inequality.
The study of the behavior of means under equal increments of their variables was initiated by Hoehn and Niven [16].Aczél and Páles [1] proved ∆ 1,s,t ≤ 1 for any s = 1.We can interpret their result as an assertion of the monotonicity of A n,t − M n,s,t as a function of t.The asymptotic behavior of t(M n,r,t − A n,t ) was studied by Brenner and Carlson [7].The same idea of [14] can be used to show that both (1.2) and (1.3) are equivalent to which holds for all t ≥ 0, In Section 3, we will study the monotonicities of (t + x n )(M n,r,t − M n,s,t ) and (t + x 1 )(M n,r,t − M n,s,t ) as functions of t for r = 1 or s = 1 and then apply the result to inequalities of the type (1.2).
The study of the behavior of means under equal increments of their variables provides us with a new approach of studying Ky Fan-type inequalities.In Section 4, we use this approach to show that some of the inequalities we have studied are actually equivalent.
The following inequality connecting three classical means (with ω i = 1/n here) is due to W. L. Wang and P. F. Wang [24] (left-hand side inequality) and Alzer et al. [5] (righthand side inequality): (1.5) The above inequality was refined in [14] and in Section 5 we will give another refinement of the above inequality.
Alzer [4] has given a counterexample to show that A α n − G α n and A α n − G α n are not comparable in general for α > 1.However, Pečarić and Alzer [22] (see also [2,Theorem 7.2] proved the following result.
Theorem 1.2.For ω i = 1/n, x n ≤ 1/2, n for α = 1/q with q = min{ω i }, a result we will establish in Section 6.A similar result is also proved there.
Let r ∈ {0, 1,...,n}; the rth symmetric function E n,r of x and its mean P n,r are defined by The usage of E n,r , E n,r , E n,r,t , P n,r , P n,r , P n,r,t is similar to the case for the power means.

Peng Gao 3553
Many of the results we will obtain for power means also have their analogues for symmetric means, which we will spend the last section to explore.For example, the result of W. L. Wang and P. F. Wang mentioned above is more general; they have shown the following. (1.8) We also note the following result of Bullen and Marcus [8].
with equality holding if and only if In Section 7, we will provide a refinement of Theorems 1.3 and 1.4 for k = 1.

Lemmas
Lemma 2.1.Let J(x) be the smallest closed interval that contains all of x i and let f (x),g(x) ∈ C 2 (J(x)) be two twice differentiable functions, then for some ξ ∈ J(x), provided that the denominator of the left-hand side is nonzero.
Lemma 2.1 and the following consequence of it are due to Mercer [17].
with equality holding if and only if and equality holds if and only if one of the following conditions holds: Moreover, when (ii) or (iii) happens, it also holds that f (x;q) = f (x t ;q) = 0.
Proof.It is routine to check f (x;q) = f (x t ;q) when t = 0 holds.So now we may assume that t > 0. Then one checks easily that ( Hence, Since x/ y > (x + t)/(y + t) when x > y > 0, t > 0, it follows from the above expression that f (x;q) ≤ f (x t ;q).The conditions for equality can be checked easily and this completes the proof.

Peng Gao 3555
Proof.It is routine to check g(x;q) = g(x t ;q) when t = 0 holds.So now we may assume that t > 0. Then one checks easily that (2.9) Hence, (2.10) Since x/ y > (x + t)/(y + t) when x > y > 0, t > 0, it follows from the above expression that g(x;q) ≤ g(x t ;q).The conditions for equality can be checked easily and this completes the proof.
Proof.We have f (q) = 2 x 1/q − y 1/q − 2 ln x 1/q x 1/q − ln y 1/q y 1/q .(2.12) So it is enough to show that u − ulnu increases for 0 < u ≤ 1 and decreases for u ≥ 1 and this is easy to check.

Some monotonicity properties
with equality holding if and only if t = 0 or Since x is arbitrary, ∆ r,s,t,α ≤ 1 is then equivalent to f (0) ≤ 0 or the second inequality below: Now ∆ r,s,t,β ≤ 1 follows from the first inequality above.This proves the first assertion and the second assertion follows similarly.
(ii) We will prove the first assertion for 0 < α < 1 and the other proofs are similar.Let We also have where the first inequality above follows from the mean value theorem and the second inequality follows from ∆ r,s,t ≤ x n /(t + x n ).Similarly, by using the mean value theorem, we get where the last inequality follows from M r n,r and adding the above two inequalities.
(v) We will prove the left-hand side inequality of (3.1) and the other proofs are similar.For 0 ≤ s < 1, let ( We want to show D n ≥ 0 here.We can assume x 1 < x 2 < ••• < x n and prove by induction that the case n = 1 is clear, so we will start with n > 1 variables assuming the inequality holds for n − 1 variables.Then where the inequality follows from ∆ 1,s,t ≤ 1.Now we consider and we have where y = (t + x n )/M n,s,t ≥ 1 and the first inequality above follows from (M n,s,t /M n,s−1,t ) 1−s ≥ 1.The last inequality above follows from the arithmetic-geometric mean inequality.
Thus s x s n and it follows ∂D n /∂x n ≥ 0 and by letting x n tend to x n−1 , we have D n ≥ D n−1 (with weights ω 1 ,...,ω n−2 ,ω n−1 + ω n ) and thus the right-hand side inequality of (3.1) holds by induction.It is easy to see that the equality holds if and only if t = 0 or For −1 ≤ s < 0, we have (3.12) The last inequality holds, since when −1 ≤ s < 0, 2 ≤ j ≤ n, we have and x j /x 1 ≥ (t + x j )/(t + x 1 ) so that Thus by a similar argument as above, we deduce that f (x 1 ) ≥ −t and ∂D n /∂x 1 ≤ 0, which further implies D n ≥ 0 with equality holding if and only if t = 0 or For 1 < s ≤ 2, it suffices to show ∂D n /∂t ≤ 0 or, equivalently, The above inequality certainly holds for s = 2, otherwise it follows from

Peng Gao 3559
We remark here that one cannot compare M n,r,t − M n,s,t and M n,r − M n,s in general.For example, let Proof.This follows from Theorem 3.1 (v) and that (1.2) and (1.4) are equivalent.
The above result was first proved by the author in [14, Theorem 3.2]; in fact it was shown there that those are the only cases (1.2) can hold for r = 1 or s = 1.Thus by noticing the equivalence of (1.2) and (1.4), we obtain the following.
) is a decreasing function of t and f (0) ≤ 0 implies the left-hand side inequality of (3.16).The proof of the right-hand side inequality of (3.16) is similar.
By a change of variables x i → 1/x n−i+1 , the left-hand side inequality of (3.16) when s = −1 gives a refinement of the left-hand side inequality of (1.2) for r = 1, s = −1.
Proof.We first show (ii)⇒(iii)⇒(i) and similarly one can show (i)⇒(iv)⇒(ii). (iii)⇒(i).Let It is easy to see that lim t→∞ f (t) = σ n /2.Thus it suffices to show f (t) is a decreasing function of t in order to prove (i).Since x is arbitrary, it suffices to have f (0 It is easy to see that lim t→∞ f (t) = 0, so it suffices to show f (t) ≥ 0 in order to prove (iii).Since x is arbitrary, it suffices to show f (0) ≥ 0. Calculation yields By a change of variables x i → 1/x n−i+1 , the right-hand side inequality of (4.1) becomes by (ii).Now we show that (i) and (v) are equivalent; similarly one can show that (i) and (vi) are equivalent.
It is easy to see that lim t→∞ f (t) = 0, so it suffices to show f (t) ≤ 0 in order to prove (i).Since x is arbitrary, it suffices to show f (0) ≤ 0, which is just (v).
Theorem 4.2.The following inequalities are equivalent: Moreover, it also holds that with equality holding if and only if x 1 = ••• = x n , which implies (3.17), and (3.17) further implies Proof.We first show that inequality (i) is equivalent to (ii).Let . By using similar arguments as in the proof of Theorem 4.1, (ii) holds if f (0) ≥ 0; by a change of variables x i → 1/x n−i+1 , one checks that f (0) ≥ 0 is equivalent to (i).
Similarly, one can show that (4.3) implies (3.17), and hence A n − H n ≤ σ n /x 1 .It now remains to show (4.3).In fact, both (4.3) and (3.17) follow from the following identity: Peng Gao 3561 We note here that it follows from (4.5) that for Upon integration of both sides above from 0 to ∞, we obtain an identity of Dinghas [12, (9.9)]: where, as given in [12, (9.8)], (4.8) We note here that (4.5) also improves Mercer's proposition [20,Proposition 4], as one checks directly.To end this section, we give a refinement of (1.2) based on the results in [20].
holds with equality holding if and only if Proof.The second inequality is [20, Proposition 1] and the first inequality follows from where the first inequality above follows from the mean value theorem and the second inequality follows from [20, Proposition 2].

It then follows by induction that
0 and this completes the proof.By letting t → ∞ in (5.1) and (5.2), we recover the following result of the author [14, Theorem 5.1], which can be regarded as a sharpening of Sierpi ński's inequality [23] for the weighted cases.
Proof.We will show the first assertion and the proof for the other one is similar.By Theorem 3.1(i), it suffices to prove the above result for α n,t ; by similar argument as in the proof of Theorem 3.1(i), it suffices to show f (0) ≤ 0, which is equivalent to A q nH 1−q n ≤ G n , and this last inequality follows from (5.7) and this completes the proof.

Some refinements of Ky Fan-type inequalities
Theorem 1.2 and Corollary 5.4 motivate the following two results.
with equality holding if and only if n = 2, q = 1/2, or Proof.We prove the right-hand side inequality of (6.1) first.By homogeneity, we may assume 0 ≤ x 1 ≤ x 2 ≤ ••• ≤ x n = 1 here and define We want to show D n ≥ 0. Let a = (a 1 ,...,a n−1 ) ∈ [0,1] n−1 be the point in which the absolute minimum of D n is reached.
We may assume a 1 ≤ a 2 ≤ ••• ≤ a n−1 and let a n = 1.If a i = a i+1 for some 1 ≤ i ≤ n − 1, by combining a i with a i+1 and ω i with ω i+1 , while noticing that increasing q will decrease the value of (1 − q)(A ) by Lemma 2.5, we can reduce the determination of the absolute minimum of D n to that of D n−1 with different weights.Thus without loss of generality, we may assume

and we obtain
∇D n a 1 ,...,a n−1 = 0 (6.3) such that a 1 ,...,a n−1 solve the equation The above equation has at most two roots (regarding A n , G n as constants), so we are reduced to the case n = 3.But if a 1 < a 2 < 1 both satisfy (6.4), we will have which is impossible since ω 1 + q ≤ 1, ω 2 + q ≤ 1 and the two equalities cannot hold at the same time.Thus if a 1 > 0, we only need to prove D 2 ≥ 0. In this case, by letting x = a 1 > 0, we get It is easy to check D 2 (1) = D 2 (1) = 0 and with equality holding if and only if x = 1 or q = 1/2.Hence, by the Taylor expansion at 1, D 2 (x) ≥ 0 with equality holding if and only if If a is a boundary point of [0,1] n−1 , then a 1 = 0 and (6.2) is reduced to We now show E n ≥ 0. Let (a 2 ,...,a n−1 ) ∈ [0,1] n−2 be the point in which the absolute minimum of E n is reached.Similar to the argument above, we may assume 0 = a 1 < a 2 < ••• < a n−1 < 1 and it is easy to show by using the method above that we only need to consider the case n = 3.When n = 3 and 0 < a 2 < 1, we have Peng Gao 3565 Using this, we get This implies that E 3 (x) takes its local maximum at a 2 , so in order to show E 3 ≥ 0, we only need to show it for the cases a 2 = 0 or a 2 = 1 and we are then back to the case n = 2.In this case, E 2 ≥ 0 is equivalent to q 1/(1−q) ≥ q/2 and g(q) = (1 − q) 1/(1−q) − q/2 ≥ 0. The first inequality follows from Lemma 2.6 and one checks that g(q) is a decreasing function of q, hence g(q) ≥ g(1/2) = 0.This now completes the proof for the right-hand side inequality of (6.1).
For the left-hand side inequality of (6.1), we may assume We want to show Again by Lemma 2.5 we may assume

and we obtain
∇F n a 2 ,...,a n = 0 (6.12) such that a 2 ,...,a n solve (6.4), which has at most two roots (regarding A n , G n as constants), so we are reduced to the case n = 3.But if 1 < a 2 < a 3 , then we will have (6.5), which is again impossible.Thus we only need to consider the case n = 2 and F 2 ≥ 0 can be proved similarly to our treatment of D 2 ≥ 0. One checks easily here F 2 (x 2 ) = 0 if and only if x 2 = 1 or q = 1/2.
So now we only need to consider the case a n → ∞.In this case, it is easy to see that if This now completes the proof for the left-hand side inequality of (6.1) with the conditions for equality readily checked.
Theorem 6.2.For 0 < q ≤ min{ω i }, σ n (6.14) with equality holding if and only if n = 2, q = 1/2, or Proof.We prove the left-hand side inequality first.By homogeneity, we may assume 0 ≤ x 1 ≤ x 2 ≤ ••• ≤ x n = 1 here and define We want to show D n ≥ 0. Let a = (a 1 ,...,a n−1 ) ∈ [0,1] n−1 be the point in which the absolute minimum of D n is reached.As in the proof of Theorem 6.1 and again using Lemma 2.5, we may assume

and we obtain
∇D n a 1 ,...,a n−1 = 0 (6.16) such that a 1 ,...,a n−1 solve the equation The above equation has at most two roots (regarding A n , G n as constants), so we are reduced to the case n = 3.But if a 1 < a 2 < 1 both satisfy (6.17), we will have which is impossible since ω 1 ≥ q, ω 2 ≥ q and the two equalities cannot hold at the same time.Thus if a 1 > 0, we only need to prove D 2 ≥ 0. In this case, set x = a 1 > 0 and consider first the case ω 1 = 1 − q, ω 2 = q.Define g(u) We first observe that by (6.17), Note for u ≥ x, q ≤ 1/3, since 0 < x < 1 and G 2 = x 1−q .As x < A 2 in our case, we then have with equality holding if and only if q = 1/2.As D 2 (1) = D 2 (1) = 0, this shows D 2 (x) ≥ 0 by considering the Taylor expansion of D 2 at 1.
Finally, we consider the case when D n reaches its absolute minimum at a with a 1 = 0. Define We show now 1, be the point in which the absolute minimum of E n is reached.Similar to the argument in the proof of Theorem 6.1, we may assume 0 = a 1 < a 2 < ••• < a n−1 < 1 and it is easy to show by using the method there that we only need to consider the case n = 3.When n = 3 and 0 < a 2 < 1, then we have Hence, So we only need to consider the case n = 2 and E 2 ≥ 0 is equivalent to g(q) = (1 − q)/2 − q 1/q ≥ 0 and (1 − q)/2 − (1 − q) 1/q ≥ 0, the second inequality follows from Lemma 2.6 and one checks that g(q) is a decreasing function of q so that g(q) ≥ g(1/2) = 0 with equality holding if and only if q = 1/2.This completes the proof for the left-hand side inequality of (6.14).
For the right-hand side inequality of (6.14), we may assume 1 (6.26) We want to show F n ≥ 0. Let a = (a 2 ,...,a n ) ∈ [1,∞) n−1 be the point in which the absolute minimum of F n is reached.
Again by Lemma 2.5 we may assume then a is an interior point of [0,1] n−1 , and we obtain ∇F n a 2 ,...,a n = 0 (6.27) such that a 2 ,...,a n solve (6.17), which has at most two roots (regarding A n , G n as constants), so we are reduced to the case n = 3.But if 1 < a 2 < a 3 , then we will have (6.18), which is again impossible.Thus we only need to consider the case n = 2 and F 2 ≥ 0 follows similarly to our treatment of D 2 ≥ 0. One checks easily here F 2 (x 2 ) = 0 if and only if x 2 = 1 or q = 1/2.So now we only need to consider the case a n → ∞.In this case it is easy to see that if q = 1/2, then n = 2 and F 2 = 0.If q < 1/2, then 1/q > 2, hence (6.28) This now completes the proof for the right-hand side inequality of (6.1) with the conditions for equality readily checked.

Results on symmetric means
Our key tool in studying the symmetric means is Lemma 2.7.We remark here that it follows from the proof of the lemma (see, e.g., [2, pages 317-318]) that for any t ≥ 0, P n,i (x t ) = P r,i (y t ).For an application of the lemma, we note the following result (see [15, Theorems 51 and 52], but be aware of the changes in notation).
We note that (7.6) is a result of Dinghas [11, page 156], originally written as Proof.Let f (t) = (x n + t) 2−r (A r n,t − P r n,r,t ).By (7.5), f is an increasing function of t and f (0) ≥ 0 gives the left-hand inequality of (7.11) and the right-hand inequality of (7.11) follows similarly.(7.12) Proof.The proof is similar to the proof of Theorem 7.3, once we note the following identity of Muirhead [21] (see also [15, Theorem 54]): where (r,i) 2 , the summation extending over all products formed from the x and of the type shown.
We note here that by taking the limit as t → ∞ in Theorem 7.5, we obtain We leave the proof of the following corollary to the reader since it is similar to the proof of Corollary 7.4.
Theorem 7.7.For t ≥ 0, n ≥ r ≥ 2, f (t;α) = P α n,1,t − P α n,r,t is a decreasing function of t for α ≤ r/(r − 1) and P α n,1,t − P α n,r,t is an increasing function for α ≥ r.In particular, for n ≥ 3, ω i = 1/n, one has n H holds with equality holding if and only if x 1 = ••• = x n when r = n and x n+1 = A n when r = n.
Proof.We use the idea in [8] and we may assume n > r.In this case, (7.23)