The proof of three power-exponential inequalities

In this paper we prove three power-exponential inequalities for positive real numbers. In particular, we conclude that this proofs give affirmatively answers to three, until now, open problems (conjectures~4.4, 2.1 and 2.2) posed by C{\^i}rtoaje in the following two works:"{\it J. Inequal. Pure Appl. Math.} 10, Article 21, 2009"and"{\it J. Nonlinear Sci. Appl.} 4:2:130-137, 2011". Moreover, we present a new proof of the inequality $a^{ra}+b^{rb}\ge a^{rb}+b^{ra}$ for all positive real numbers $a$ and $b$ and $r\in [0,e]$. In addition, three new conjectures are presented.


Introduction
The power-exponential functions have useful applications in mathematical analysis and in other theories like Statistics [1], Biology [2,3], Optimization [4], Ordinary Differential equations [5] and Probability [6]. In the recent years there is a intensive research in this area, see for instance [7,8,9,10,11,12,13,14,15,16,17] and the recent overview on general mathematical inequalities done by Cerone and Dragomir [18]. Some problems look like very simple but, are difficult to solve. For instance, we have the following two classical problems: find the solution of the equation ze z = a and the basic problem of comparing a b and b a for all positive real numbers a and b. The first problem is perhaps one of the most ancient and useful problems concerning to power-exponential functions, see for instance [19,20,21]. It was introduced by Lambert in [22] and have been studied by recognized mathematicians like Euler, Pólya, Szegö and Knuth, see [23,24,25]. The solution to the problem have been inspirited the definition of the well-known W -Lambert function, see [26]. For the solution to the second problem, see the discussion given in [27,28] and more recently in [16]. Moreover, in spite of its algebraic simplicity, booth problems are the central topic of a large number of research papers in the last years (see [7,11,13] and references therein). In particular, in this paper, we are interested in some inequalities conjectured by Cîrtoaje in [12,29], which are very close to the second problem. To be more specific, we start by recalling that in [30] was introduced and probed the following assertion: the inequality a a + b b ≥ a b + b a holds for all positive real numbers less than or equal to 1. After that, Cîrtoaje [12] introduce, prove and conjecture several results about inequalities for power-exponential functions. In particular, in [12], was established that the inequality Conjecture 4.7. If a and b are nonnegative real numbers such that a + b = 2, then Conjecture 4.8. If a and b are nonnegative real numbers such that a + b = 1, then a 2b + b 2a ≤ 1. Afterwards, the analysis of (1.1) was completed by Manyama in [14]. Thereafter, of the Cîrtoaje conjectures, the milestones of the history are the works of Coronel and Huancas [13], Matejíčka [31], Yin-Li [9] and Hisasue [10] (see also the work of Cîrtoaje [29]), where they proved the conjectures 4.3, 4.6, 4.7 and 4.8, respectively. Here, we should be comment that the proof of conjecture 4.4 is still open. Subsequently, in 2011 Cîrtoaje introduce a new proof of (1.1) and present the following three new conjectures Recently, Miyagi and Nishizawa [7] has been proved the Conjecture 5. The main contribution of the present paper is the development of the proof of the following three theorems:   holds.
Note that the conjectures 4.4, 2.1 and 2.2 are solved by the Theorems 1.2, 1.3, 1.4, respectively. Moreover, we develop a proof of Theorem 1.1 which is an alternative proof of (1.1) for all positive real numbers a, b and r ∈ [0, e], which is distinct to the existing proofs given in [14,29]. The rest of the paper is organized in two sections: In section 2 we present the proofs of Theorems 1.1, 1.2, 1.3 and 1.4 and in section 3 we present some remarks and three new conjectures.

Proofs of main results
In this section we present the proofs of Theorems 1.1, 1.2, 1.3, 1.4. Firstly, we recall a result of [13]. Then, we present the corresponding proofs.

A preliminar result.
For completeness and self-contained structure of the proofs of Theorems 1.1 and 1.2, we need the following result of [13].
Proposition 2.1. Consider s ∈ R + with s = 1, m ∈ R + and f, g : R + → R defined as follows Then, the following properties are satisfied (iv) g is continuous on R + ∪ {0} and strictly increasing on R + . Furthermore y = 1 is a horizontal asymptote of y = g(t).

2.2.
Proof of Theorem 1.1. Without loss of generality, we assume that a > b. Indeed, we follow the proof (1.1) by application of Proposition 2.1 with t = a rb , γ = b rb and s = a/b. Indeed, we distinguish three cases (a 2 ) Case a > b > 1 t > γ > 1 and s > 1 . By Proposition 2.1-(iv), we note that g(s) < 1. Then, by the strictly increasing behavior of f (Proposition 2.1-(ii)) we deduce the inequality since: , 1], we follow the inequality by almost identical arguments to that used before in (i), since t > 1 ≥ γ ≥ g(s) and s > 1.
we deduce that The function h is concave and has a maximum at (1/e, r/e). Thus, we deduce that Secondary, by the Napier's inequality [32]: which implies γ > g(s). The proof of this case is completed by application of Proposition 2.1-(ii). Hence, by (a 2 ), (b 2 ) and (c 2 ) we follow that Theorem 1.1 is valid.
2.3. Proof of Theorem 1.2. The proof of this theorem is again developed by application of Proposition 2.1. Firstly, we recall the notation of [13]: The family is a set partition of R 3 + . Now, with this notation, we subdivide the proof in three parts: This special case is a direct consequence of Theorem 1.1.
we apply the Theorem 1.1 and Proposition 2.1 as follows. We select t = a rb , γ = c rb and s = a/b, the monotonic behavior and properties of function f , defined on Proposition 2.1, implies that since t > γ, t > 1 and s > 1. Indeed, the corresponding proof of (2.3) needs the distinction of two cases: c ≥ 1 and c < 1. If c ≥ 1, then γ > 1 and γ ∈ ]g(s), ∞[, so f is strictly increasing and t > γ implies (2.3). For c < 1, we note that γ < 1 and −γ s + γ ≥ 0 since s > 1 and 1 ∈]g(s), ∞[, then the assumption t > 1 implies that (2.3) is again true for this subcase. Moreover, for (a, b, c) ∈ E + a ⊂ R 3 + , by Theorem 1.1, we recall that the inequality The proof for (a, b, c) ∈ E + b ∪ E + c is similar to the case (a, b, c) ∈ E + a and we omit the details. However, we comment that for (a, b, c) ∈ E + b we choose t = b rc , γ = c 2c and s = b/c and for (a, b, c) ∈ E + c we select t = c ra , γ = b ra and s = c/a.
Without loss of generality, we assume that (a, b, c) ∈ E − a is such that c < b < a, since the proof for b < c < a is similar. We note that Ω = [0, e] × [0, 1] can be partitioned in the two sets and Now, we continue the proof by distinguish the following two subcases: (r, c) ∈ Ω 1 and (r, c) ∈ Ω 2 . For the subcase (r, c) ∈ Ω 1 , we apply the function f given on Proposition 2.1 with t = b rc , γ = c rc and s = a/c to prove b ra + c rc > b rc + c ra for 0 < c < b < a < 1 and (r, c) ∈ Ω 1 . (2.6) Now, from (2.6), we note that which implies (2.5) by application of Proposition 2.1-(ii), since t > γ > g(s) and f is increasing on ]g(s), ∞[. For the subcase (r, c) ∈ Ω 2 , we apply the function f given on Proposition 2.1 with t = b rc , γ = c rc and s = a/c to prove b ra + c rc > b rc + c ra for 0 < c < b < a < 1 and (r, c) ∈ Ω 2 . (2.8) We note that the inequality c rc > c r−1 holds true for all (r, c) ∈ Ω 2 . Now, in order to deduce that γ > g(s) is suficiently to prove that c r−1 > g(s). Indeed, the function q : [c, 1] → R defined as follows q(z) = c (1−r)z z c − c c+c(1−r) has the following properties: Then, we deduce that q(z) ≥ 0 for all z ∈ [c, 1]. In particular, for z = a ∈ [c, 1], we deduce that c (1−r)a a c − c c+c(1−r) ≥ 0, which implies the following sequence of implications . Thus (2.8) holds true.
From (2.5) and (2.8), we deduce that b ra + c rc > b rc + c ra for 0 < c < b < a < 1 and r ∈ [0, e]. (2.9) Hence, to complete the proof for 0 < c < b < a < 1, we add the inequality (2.9) with a ra + b rb > a rb + b ra for r ∈ [0, e], which is true by Theorem 1.1.
Then we prove that H(x) > 0 for all x ∈]0, 1], which naturally implies the inequality 2 √ a ra b rb ≥ a rb + b ra for x = a. Indeed, we prove that the function H has a global minimum at x = b. The fact that in x = b there is a local minimum of H, follows by noticing that H ′ (b) = 0 and H ′′ (b) > 0, since Meanwhile, the property that b is a global minimum of H can be proved by rewriting H ′ as the difference of two functions and by analyzing the sign of H ′ using some properties of this new functions. Indeed, to be more specific, we note that where the functions K and Q are defined as follows The functions K and Q have the following properties We note that the sets Λ i , i = 1, 2, 3, are not empty since for instance ]0, 1[×[1/b, e] ⊂ Λ 1 for all b ∈]0, 1], {1}×]0, 1[⊂ Λ 2 and ]0, 1[×{1} ⊂ Λ 3 . Moreover, we note that rb > 1 implies that (b, r) ∈ Λ 1 and naturally Λ 2 ∪ Λ 3 is a subset of ]0, 1]×]0, 1/b[. The uniqueness of c can be deduced by noticing that the solution of Q ′ (x) = 0 is equivalent to the intersection of the following two monotone functions S(x) = rb rx (ln b) 2 and J( From (K 1 ) and (Q 1 ) we deduce the uniqueness of b ∈]0, 1] such that Q(b) = K(b) or equivalently H ′ (b) = 0. Now, from (K 2 ) and (Q 2 ), we note that Q(0 + ) > K(0 + ) for all (r, b) ∈ Λ since K(0 + ) = −∞. Then, H ′ (x) < 0 for all x ∈]0, b[. Additionally, from (K 2 ) and (Q 2 ), we observe that Q(1) < K(1). This fact is a consequence of that the function F (w, r) = √ w rw − w r ln(w) − w is strictly decreasing in r, since F r (w, r) = ln(w) (r/2) √ w rw − w r ln(w) < 0. Consequently, for r < e we have that F (w, r) > F (w, e) = √ w ew − w e ln(w) − w > 0 for all w ∈]0, 1]. Hence, for w = b we get that F (b, r) > 0 or Q(1) < K(1), which implies that H ′ (x) > 0 for all x ∈]b, 1]. Thus, b is a global minimum of H. Therefore, H(x) ≥ H(b) = 0 for all x ∈]0, 1] and in particular for x = a.

2.5.
Proof of Theorem 1.4. The proof follows by the fact that the function P :]0, 1] n−1 → R defined the following correspondence rule has a global minimum at (z 1 , . . . , z n−1 ) = (x n , . . . , x n ). Indeed, by simplicity of notation we develop the details of the proof for n = 3 and with (x 1 , x 2 , x 3 ) = (a, b, c). Note that, in this case for an arbitrary c ∈]0, 1], the function P :]0, 1] 2 → R has the following form Then, we have that An evaluation at (c, c) implies that P x (c, c) = P y (c, c) = 0, P xx (c, c) = P yy (c, c) = c 3c−1 − 6c(ln(c)) 2 + 4 , Now, defining P 1 (w) = −6w(ln(w)) 2 + 4 and P 2 (w) = 27w 2 (ln w) 4 − 24w(ln w) 2 + 4, we observe that P xx (c, c) = c 3c−1 P 1 (c) and P xx (c, c)P yy (c, c) − P xy (c, c)P xy (c, c) = c 2(3c−1) P 2 (c). Then, the Hessian matrix asociated to P at (c, c) is positive semidefinite since both functions, P 1 and P 2 , are positive on ]0, 1] or equivalently the function P has a local minimum at (c, c). Now, we deduce that (c, c) is the global minimum since we can prove that (c, c) is the unique solution of (P x , P y ) = (0, 0). Indeed, assuming that there is (x, y) with x = y = c such that P x (x, y) = P y (x, y) = 0, we can deduce a contradiction. Note that since the inequality ln(r) > (r − 1)/r holds for all r > 0 and r = 1 (see for instance [27]). Then, x = y, which is a contradiction with the assumption that x = y. Thus, we have (c, c) is a global minimum of the function P or equivalently P (x, y) ≥ P (c, c) = 0 for all (x, y) ∈]0, 1] 2 , which implies the desired inequality for (x, y) = (a, b).

Aditional remarks on posible generalizations
In this section we present the posible extensions of Theorems 1.1, 1.2 and 1.3 to a sequence of positive real numbers. We note that the natural generalizations of (1.2) and (1.3) are given by respectively. We present a partial proof of (3.1) (see Lemma 3.1, below) and leaves as a conjecture the proof of (3.2). Proof. Before, of the start the proof, we notice that the function Υ(x, y) = x a/b − x − y a/b + y defined from R 2 + → R and for a > b is concave and Υ(0, 0) = Υ(1, 0) = Υ(0, 1) = Υ(1, 1) = 0. Then Υ(x, y) ≥ 0 for all (x, y) ∈ [0, 1] × [0, 1]. Similarly, the function Υ s (w, z) = Υ(z, w) for a < b is concave and Υ s (w, z) ≥ 0 for all (w, z) ∈ [0, 1] × [0, 1]. Now, we proceed by induction on n. Let us assume that the theorem is valid for a sequence of positive numbers (x 1 , . . . , x k ) for all k < n. We note that 3) The terms K 1 and K 2 are positive by the inductive hypothesis. Meanwhile, the term K 3 is positive by the coancavity of the functions Υ and Υ s . Note that a = x n−1 and b = x n and K 3 = Υ(x rx1 n , x rx1 n−1 ) or K 3 = Υ s (x rx1 n−1 , x rx1 n ), depending if x n−1 > x 1 or x n−1 < x 1 , respectively. Then, by (3.3) we follow that the Lemma is valid.