Properties of the power-mean and their applications

: Suppose w , v > 0, w (cid:44) v and A u ( w , v ) is the u -order power mean (PM) of w and v . In this paper, we completely describe the convexity of u (cid:55)→ A u ( w , v ) on R and s (cid:55)→ A u ( s ) ( w , v ) with u ( s ) = (ln 2) / ln (1 / s ) on (0 , ∞ ). These yield some new inequalities for PMs, and give an answer to an open problem


Introduction
A function M : R 2 + → R is called a bivariate mean (BM) if for all w, v > 0 is valid.It is said to be homogeneous (of degree one) if for all λ, w, v > 0 is valid.If a BM M is differentiable on R 2 + , then the function M u : R 2 + → R defined by M u (w, v) = M 1/u (w u , v u ) if u 0 and M 0 (w, v) = w M x (1,1) v M y (1,1) , ( is called "u-order M mean", where M x (x, y), M y (x, y) are the first-order partial derivatives in regard to the first and second components of M(x, y), respectively (see [1]).For example, the arithmetic mean (AM), logarithmic mean (LM) and identric mean (IM) are given by , respectively, then if u 0 and I 0 (w, v) = √ wv (1.4) are u-order AM, u-order LM and u-order IM, respectively.As usual, the u-order AM is still called u-order PM.Correspondingly, since the form of M u is similar to PM A u , it is also known simply as "power-type mean".More general means than power-type mean including Stolarsky means, Gini means, and two-parameters functions, etc., which can be seen in [2][3][4][5][6][7].
In this paper, we are interested in the properties of the PM A u .As is well-known that u → A u (w, v) is increasing on R (see [5]).The log-convexity of u → A u (w, v), L u (w, v) and I u (w, v) is a direct consequence of [9,Conclusion 1. 1)] when q = 0, that is, Theorem 1.The functions u → A u (w, v), L u (w, v) and I u (w, v) are log-convex on (−∞, 0) and logconcave on (0, ∞).
The log-convexity of the function u → A u (w, v) was reproved in [19] by Begea, Bukor and Tóhb.The authors proposed an open problem on the convexity of the function u → A u (w, v): Problem 1 was proven by Matejíčka in [20].In 2016, Raïsouli and Sándor [16,Problem 1] proposed the following problem.
Clearly, this problem is partly related to the convexity of u → A u (w, v).Motivated by Problem 2, the main purpose of this paper is to investigate completely the convexity of u → A u (w, v) on R and s → A u(s) (w, v) with u (s) = (ln 2) / ln (1/s) on (0, ∞).As applications, some new inequalities for power means are established, and an answer to Problem 2 is given.Final, three problems on the convexity of certain power-type means and inequalities are proposed.
It should be noted that a homogeneous BM can be represented by the exponential functions.If M (x, y) is a HM of positive arguments x and y, then M (x, y) can be represented as where t = (1/2) ln (x/y).Further, if M (x, y) is symmetric, then M (x, y) can be expressed in terms of hyperbolic functions (see [18,Lemma 3]).For example, in view of symmetry, we suppose v > w > 0.
Then we find t = (1/2) ln (v/w) > 0. Thus the PM A u (w, v), u-order LM L u (w, v) and u-order IM I u (w, v) can be represented as The first result of the paper is the following theorem.
where and in what follows the symbols "∪" and "∩" denote the given function are convex and concave, "∩∪" and "∪∩" denote the given function are "concave then convex" and "convex then concave", respectively.
The second and third results of the paper are the following theorems.
Remark 3. By Theorems 3 and 4, the function s → A u(s) (w, v) has the following (log-) convexity:

Tools
To prove the lemmas listed in Sections 3-5, we need two tools.The first is the so-called L'Hospital Monotone Rule (LMR), which appeared in [21] (see also [22]).
Proposition 1. Suppose −∞ ≤ a < b ≤ ∞, φ and ψ are differentiable functions on (a, b).Suppose also the derivative ψ is nonzero and does not change sign on (a, b), and φ(a Before stating the second tool, we present first an important function H φ,ψ .Assume that φ and ψ are differentiable functions on (a, b) with ψ 0, where −∞ ≤ a < b ≤ ∞.It was introduced by Yang in [23, Eq (2.1)] that which we call Yang's H-function.This function has some good properties, see [23, Properties 1 and 2], and plays an important role in the proof of a monotonicity criterion for the quotient of two functions, see for example, [24][25][26][27][28].
To study the monotonicity of the ratio φ/ψ on (a, b), Yang [23, Property 1] presented two identities in term of H φ,ψ , which state that, if φ and ψ are twice differentiable with ψψ 0 on (a, b), then (2.3)

Proof of Theorem 2
In order to prove Theorem 2, we need the following lemma.
Proof.Differentiation yields Then .
Proof of Theorem 2. Differentiation yields Letting ut = x and simplifying give where h 1 (x) and g 1 (x) are given in Lemma 1.Since h 1 (x) and g 1 (x) are even on (−∞, ∞) and From Lemma 1 we find We conclude thus that F t (u) > (<) 0 for all t > 0 if and only if When ln Since for x ∈ (0, ∞), the function h 1 (x) is strictly decreasing, the inverse of h 1 exists and so is h −1 1 .Solving the equation (3.3) for t yields Noting that 1/u and h −1 1 (u) are both positive and decreasing, so is t = T 1 (u).This implies u = T −1 1 (t) exists and strictly decreasing on (0, ∞).It then follows that where u 1 = T −1 1 (t 1 ).We thus arrive at that , which completes the proof.

Proof of Theorem 3
Lemma 2. The function is strictly decreasing from (0, ∞) onto 0, ln Proof.We write where It is easy to check that Differentiation yields where Then , where g 4 (x) 0. As shown in the proof of Lemma 1, f 4 (x) /g 4 (x) is strictly decreasing on (0, ∞).
Taking into account Cases 1 and 2 as well the continuity of the function g 3 (x) at x = x 2 , we conclude that h 2 = f 3 /g 3 is strictly decreasing on (0, ∞).An easy calculation yields h 2 (0) = ln √ 2 and h 2 (∞) = 0, and the proof is completed.Now we shall prove Theorem 3.
Letting ut = x and simplifying give where We conclude thus that G t (s) ≥ (≤) 0 for all t > 0 if and only if Since the function h 2 (x) , (x > 0) is strictly decreasing, the inverse of h 2 exists and so is h −1 2 .Solving the Eq (4.2) for t yields Because that 1/u and h −1 2 (u) are both positive and strictly decreasing, so is t = T 2 (u).This implies u = T −1 2 (t) exists and strictly decreasing on (0, ∞).It then follows that where We thus deduce that G t (s) < 0 for u ∈ u 2 , ln √ 2 and G t (s) > 0 for u ∈ (0, u 2 ).Due to u = (ln 2) / ln (1/s), it follows that G t (s) < 0 on u ∈ s * 2 , e −2 and G t (s) > 0 on 0, s * 2 , where s * 2 = 2 −1/u 2 .This completes the proof.

Proof of Theorem 4
Lemma 3. The function Proof.As shown in Lemmas 1 and 2, x tanh x − ln cosh x = f 2 (x) and By Proposition 1 we deduce that g 3 (x) / f 2 (x) is strictly decreasing on (0, ∞), which, due to which completes the proof.
Based on Lemma 3, we now check Theorem 4.
Proof of Theorem 4. Differentiation yields Letting ut = x and simplifying lead to We conclude thus that [ln G t (s)] ≤ 0 for all t > 0 if and only if u ≤ ln √ 2, which, by the relation u = (ln 2) / ln (1/s), implies that 0 < s ≤ e −2 or s > 1.This completes the proof.

Several new inequalities
Using Theorems 2 and 4, we get the following corollary.
Using Theorems 1 and 3, we obtain the following corollary.
Corollary 2. Suppose w, v > 0, w v.If ln √ 2 ≤ p < r < q, then the double inequality holds, where α 0 and β 0 are given in (6.2) are the best constants.The double inequality (6.5) is reversed if p < r < q < 0 with the best constants α 0 and β 0 .
Without loss of generality, we suppose that 0 for max {p, q, r} < 0 or min {p, q, r} > 0, α 0 and β 0 are the best.This completes the proof.
Similarly, by means of Theorems 1 and 4 we can prove the following corollary, all the details of proof are omitted here.

By means of Corollaries 1 and 2, we have
Corollary 4. Suppose p, q, r ∈ R, p < r < q. (i) If p ≥ 1/2, then for w, v > 0, w v the double mean-inequality is valid if and only if (ii) If q < 0, then for w, v > 0, w v the double inequality (6.8) is reversed if and only if α ≥ α 0 and β ≤ β 0 .
(ii) The second assertion of this theorem can be proven in a similar way.This completes the proof.

Conclusions
In this paper, we completely described the convexity of u → A u (w, v) on R and s → A u(s) (w, v), ln A u(s) (w, v) with u (s) = (ln 2) / ln (1/s) on (0, ∞) by using two tools.From which we obtained several new sharp inequalities involving the power means (Corollaries 1-4), where Corollary 4 gives an answer to Problem 2.Moreover, we gave another new proof of Problem 1.
Final inspired by Theorems 1-4, we propose the following problem.
Problem 3.For all w, v > 0, w v, determine the best p ∈ R such that the functions p → L p (w, v), I p (w, v) are convex or concave .
The second problem is inspired by Corollary 3 and Problem 2.