On the Weighted AM–GM Inequality and Refined Inequalities Between Arithmetic Functions

1. Introduction

Let n be a positive integer. Let $\phi \left(n\right)$, $\psi \left(n\right)$ and $\sigma \left(n\right)$ denote the Euler totient function, Dedekind function and sum–of–divisors function respectively. For $n>1$, we have

$$\phi \left(n\right)=n\prod _{p⩽n}{\displaystyle \frac{p-1}{p}}\mathrm{and}\psi \left(n\right)=n\prod _{p⩽n}{\displaystyle \frac{p+1}{p}}.$$

(1)

These arithmetic functions satisfy many important properties. For example, the following inequality is well–known:

$$\phi \left(n\right)⩽\psi \left(n\right)⩽\sigma \left(n\right).$$

(2)

In this paper we are looking for bounds for quantities

$$\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}\mathrm{and}\phi {\left(n\right)}^{\psi \left(n\right)}\psi {\left(n\right)}^{\phi \left(n\right)}.$$

In 2011, Atanassov [1] first obtained a lower bound for $\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}$. He showed that for any $n>1$, we have

$$\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}>n^{2n}.$$

(3)

In 2013, Kannan and Srikanth [2] sharpened (3) by showing that

$$\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}>n^{\phi \left(n\right)+\psi \left(n\right)}.$$

(4)

Finally, Sándor and Atanassov [3] in 2019 proved the following refined estimates using the weighted AM–GM inequality.

$$\begin{array}{cc}\hfill n^{\phi \left(n\right)+\psi \left(n\right)}<{\left({\displaystyle \frac{\phi \left(n\right)+\psi \left(n\right)}{2}}\right)}^{\phi \left(n\right)+\psi \left(n\right)}<& \phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}<{\left({\displaystyle \frac{\phi {\left(n\right)}^2+\psi {\left(n\right)}^2}{2}}\right)}^{{\displaystyle \frac{\phi \left(n\right)+\psi \left(n\right)}{2}}}\hfill \end{array}$$

$$\begin{array}{cc}\hfill <& \psi {\left(n\right)}^{\phi \left(n\right)+\psi \left(n\right)},\hfill \\ \hfill {\left({\displaystyle \frac{\phi \left(n\right)\psi \left(n\right)\left(\phi \right(n)+\psi (n\left)\right)}{\phi {\left(n\right)}^2+\psi {\left(n\right)}^2}}\right)}^{\phi \left(n\right)+\psi \left(n\right)}<& \phi {\left(n\right)}^{\psi \left(n\right)}\psi {\left(n\right)}^{\phi \left(n\right)}<{\left({\displaystyle \frac{2\phi \left(n\right)\psi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)}\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill <& {\left(\phi \left(n\right)\psi \left(n\right)\right)}^{{\displaystyle \frac{\phi \left(n\right)+\psi \left(n\right)}{2}}}<n^{\phi \left(n\right)+\psi \left(n\right)}.\hfill \end{array}$$

(6)

For other types of inequalities between arithmetic functions, we refer the readers to [4] and its references. In this paper, we shall use some refined inequalities to improve the upper bounds proved by Sándor and Atanassov [3].

Theorem 1.

For any integer $n>1$, we have the following inequalities:

$$\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}⩽{\left({\displaystyle \frac{\phi {\left(n\right)}^3+4\phi {\left(n\right)}^2\psi \left(n\right)+\phi \left(n\right)\psi {\left(n\right)}^2+2\psi {\left(n\right)}^3}{2{(\phi \left(n\right)+\psi \left(n\right))}^2}}\right)}^{\phi \left(n\right)+\psi \left(n\right)},$$

(A1)

$$\begin{array}{cc}\hfill \phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}⩽& \left({\displaystyle \frac{2\phi {\left(n\right)}^2\psi \left(n\right)+2\psi {\left(n\right)}^3-\phi \left(n\right){\left(\phi \left(n\right)-\phi {\left(n\right)}^{\frac{\phi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}-\psi {\left(n\right)}^{\frac{\psi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^2}{2\psi \left(n\right)\left(\phi \right(n)+\psi (n\left)\right)}}\right.\hfill \\ \hfill & {\left.{\displaystyle \frac{-\psi \left(n\right){\left(\psi \left(n\right)-\phi {\left(n\right)}^{\frac{\phi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}-\psi {\left(n\right)}^{\frac{\psi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^2}{2\psi \left(n\right)\left(\phi \right(n)+\psi (n\left)\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)},\hfill \end{array}$$

(B1)

$$\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}⩽{\left({\displaystyle \frac{\phi {\left(n\right)}^{\frac{3}{2}}+\psi {\left(n\right)}^{\frac{3}{2}}}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^{2\left(\phi \right(n)+\psi (n\left)\right)},$$

(C1)

$$\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}⩽{\left({\displaystyle \frac{2\phi {\left(n\right)}^{\frac{3}{2}}\psi {\left(n\right)}^{\frac{1}{2}}-\phi \left(n\right)\psi \left(n\right)+\psi {\left(n\right)}^2}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)},$$

(D1)

$$\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}⩽{\left({\displaystyle \frac{\phi {\left(n\right)}^2+\psi {\left(n\right)}^2}{(\phi \left(n\right)+\psi \left(n\right))exp\left(2-2\frac{\phi {\left(n\right)}^{\frac{3}{2}}+\psi {\left(n\right)}^{\frac{3}{2}}}{{\left(\left(\phi {\left(n\right)}^2+\psi {\left(n\right)}^2\right)(\phi \left(n\right)+\psi \left(n\right))\right)}^{\frac{1}{2}}}\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)},$$

(E1)

$$\phi {\left(n\right)}^{\phi \left(n\right)}\psi {\left(n\right)}^{\psi \left(n\right)}⩽{\left({\displaystyle \frac{\phi {\left(n\right)}^2+\psi {\left(n\right)}^2-\frac{3\phi \left(n\right)\psi \left(n\right){(\psi \left(n\right)-\phi \left(n\right))}^2}{2\phi {\left(n\right)}^2+8\phi \left(n\right)\psi \left(n\right)+2\psi {\left(n\right)}^2}}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)},$$

(F1)

$$\phi {\left(n\right)}^{\psi \left(n\right)}\psi {\left(n\right)}^{\phi \left(n\right)}⩽{\left({\displaystyle \frac{-\phi {\left(n\right)}^3+6\phi {\left(n\right)}^2\psi \left(n\right)+3\phi \left(n\right)\psi {\left(n\right)}^2}{2{(\phi \left(n\right)+\psi \left(n\right))}^2}}\right)}^{\phi \left(n\right)+\psi \left(n\right)},$$

(A2)

$$\begin{array}{cc}\hfill \phi {\left(n\right)}^{\psi \left(n\right)}\psi {\left(n\right)}^{\phi \left(n\right)}⩽& \left({\displaystyle \frac{4\phi \left(n\right)\psi {\left(n\right)}^2-\phi \left(n\right){\left(\psi \left(n\right)-\psi {\left(n\right)}^{\frac{\phi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}-\phi {\left(n\right)}^{\frac{\psi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^2}{2\psi \left(n\right)\left(\phi \right(n)+\psi (n\left)\right)}}\right.\hfill \\ \hfill & {\left.{\displaystyle \frac{-\psi \left(n\right){\left(\phi \left(n\right)-\psi {\left(n\right)}^{\frac{\phi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}-\phi {\left(n\right)}^{\frac{\psi \left(n\right)}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^2}{2\psi \left(n\right)\left(\phi \right(n)+\psi (n\left)\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)},\hfill \end{array}$$

(B2)

$$\phi {\left(n\right)}^{\psi \left(n\right)}\psi {\left(n\right)}^{\phi \left(n\right)}⩽{\left({\displaystyle \frac{\phi {\left(n\right)}^{\frac{1}{2}}\psi {\left(n\right)}^{\frac{1}{2}}\left(\phi {\left(n\right)}^{\frac{1}{2}}+\psi {\left(n\right)}^{\frac{1}{2}}\right)}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^{2\left(\phi \right(n)+\psi (n\left)\right)},$$

(C2)

$$\phi {\left(n\right)}^{\psi \left(n\right)}\psi {\left(n\right)}^{\phi \left(n\right)}⩽{\left({\displaystyle \frac{2\phi {\left(n\right)}^{\frac{3}{2}}\psi {\left(n\right)}^{\frac{1}{2}}+\phi \left(n\right)\psi \left(n\right)-\phi {\left(n\right)}^2}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)},$$

(D2)

$$\phi {\left(n\right)}^{\psi \left(n\right)}\psi {\left(n\right)}^{\phi \left(n\right)}⩽{\left({\displaystyle \frac{2\phi \left(n\right)\psi \left(n\right)}{(\phi \left(n\right)+\psi \left(n\right))exp\left(2-2\frac{\phi {\left(n\right)}^{\frac{1}{2}}\psi {\left(n\right)}^{\frac{1}{2}}\left(\phi {\left(n\right)}^{\frac{1}{2}}+\psi {\left(n\right)}^{\frac{1}{2}}\right)}{{\left(2\phi \left(n\right)\psi \left(n\right)\left(\phi \right(n)+\psi (n\left)\right)\right)}^{\frac{1}{2}}}\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)}$$

(E2)

and

$$\phi {\left(n\right)}^{\psi \left(n\right)}\psi {\left(n\right)}^{\phi \left(n\right)}⩽{\left({\displaystyle \frac{2\phi \left(n\right)\psi \left(n\right)-\frac{3\phi \left(n\right)\psi \left(n\right){(\psi \left(n\right)-\phi \left(n\right))}^2}{4\left(\phi {\left(n\right)}^2+\phi \left(n\right)\psi \left(n\right)+\psi {\left(n\right)}^2\right)}}{\phi \left(n\right)+\psi \left(n\right)}}\right)}^{\phi \left(n\right)+\psi \left(n\right)}.$$

(F2)

All inequalities above may be replaced with function $\sigma \left(n\right)$ instead of function $\psi \left(n\right)$.

Note that we have

$$\left(\mathbf{B}\mathbf{1}\right)\Rightarrow \left(\mathbf{A}\mathbf{1}\right),\left(\mathbf{B}\mathbf{2}\right)\Rightarrow \left(\mathbf{A}\mathbf{2}\right),$$

$$\left(\mathbf{D}\mathbf{1}\right)\Rightarrow \left(\mathbf{C}\mathbf{1}\right),\left(\mathbf{D}\mathbf{2}\right)\Rightarrow \left(\mathbf{C}\mathbf{2}\right),$$

where $\mathbf{X}\Rightarrow \mathbf{Y}$ means that X implies Y.

2. Refinements of the Weighted AM–GM Inequality

In this section we shall list several variants of the weighted AM–GM inequality. First, recall that the classical weighted AM–GM inequality states that

Lemma 1.

If real numbers ${\alpha }_1,\dots ,{\alpha }_n>0$ satisfy ${\alpha }_1+\dots +{\alpha }_n=1$, then for $x_1,\dots ,x_n⩾0$, we have

$$x_1^{{\alpha }_1}\dots x_n^{{\alpha }_n}⩽{\alpha }_1{x}_1+\dots +{\alpha }_n{x}_n.$$

Moreover, for $n=2$ this becomes

$${\displaystyle \frac{1}{\frac{{\alpha }_1}{x_1}+\frac{{\alpha }_2}{x_2}}}⩽x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\alpha }_1{x}_1+{\alpha }_2{x}_2.$$

In Lemma 1, put $x_1=a$, $x_2=b$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$ we get

Lemma 2

([3], Proposition 1). For any $a,b>0$, we have

$${\left({\displaystyle \frac{a+b}{2}}\right)}^{a+b}⩽a^a{b}^b⩽{\left({\displaystyle \frac{a^2+b^2}{a+b}}\right)}^{a+b}.$$

Again, put $x_1=b$, $x_2=a$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$ we get

Lemma 3

([3], Proposition 2). For any $a,b>0$, we have

$${\left({\displaystyle \frac{ab(a+b)}{a^2+b^2}}\right)}^{a+b}⩽a^b{b}^a⩽{\left({\displaystyle \frac{2ab}{a+b}}\right)}^{a+b}.$$

We remark that Sándor and Atanassov [3] used Lemmas 1–3 to prove their bounds.

Next, we mention some results that yield improvements on Lemma 1. We will use them to prove our Theorem 1 in the next section.

Lemma 4

([5], Theorem, [6], Remark 3). Let $n⩾2$, $x_1,\dots ,x_n⩾0$ and ${\alpha }_1,\dots ,{\alpha }_n>0$. Suppose that ${\alpha }_1+\dots +{\alpha }_n=1$. Then we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2max(x_1,\dots ,x_n)}}\sum _{1⩽i⩽n}{\alpha }_i{\left(x_i-\sum _{1⩽j⩽n}{\alpha }_j{x}_j\right)}^2\hfill \\ \hfill ⩽& \sum _{1⩽i⩽n}{\alpha }_i{x}_i-\prod _{1⩽i⩽n}{x}_i^{{\alpha }_i}\hfill \\ \hfill ⩽& {\displaystyle \frac{1}{2min(x_1,\dots ,x_n)}}\sum _{1⩽i⩽n}{\alpha }_i{\left(x_i-\sum _{1⩽j⩽n}{\alpha }_j{x}_j\right)}^2.\hfill \end{array}$$

Moreover, for $n=2$ this becomes

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2max(x_1,x_2)}}\left({\alpha }_1{\left(x_1-{\alpha }_1{x}_1-{\alpha }_2{x}_2\right)}^2+{\alpha }_2{\left(x_2-{\alpha }_1{x}_1-{\alpha }_2{x}_2\right)}^2\right)\hfill \\ \hfill ⩽& \left({\alpha }_1{x}_1+{\alpha }_2{x}_2-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right)\hfill \\ \hfill ⩽& {\displaystyle \frac{1}{2min(x_1,x_2)}}\left({\alpha }_1{\left(x_1-{\alpha }_1{x}_1-{\alpha }_2{x}_2\right)}^2+{\alpha }_2{\left(x_2-{\alpha }_1{x}_1-{\alpha }_2{x}_2\right)}^2\right).\hfill \end{array}$$

Lemma 5

([7], Theorem).  Let $n⩾2$, $x_1,\dots ,x_n⩾0$ and ${\alpha }_1,\dots ,{\alpha }_n>0$. Suppose that ${\alpha }_1+\dots +{\alpha }_n=1$. Then we have

$${\displaystyle \frac{1}{2max(x_1,\dots ,x_n)}}\sum _{1⩽i⩽n}{\alpha }_i{\left(x_i-\prod _{1⩽j⩽n}{x}_j^{{\alpha }_j}\right)}^2⩽\sum _{1⩽i⩽n}{\alpha }_i{x}_i-\prod _{1⩽i⩽n}{x}_i^{{\alpha }_i}.$$

Moreover, for $n=2$ this becomes

$${\displaystyle \frac{1}{2max(x_1,x_2)}}\left({\alpha }_1{\left(x_1-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right)}^2+{\alpha }_2{\left(x_2-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right)}^2\right)⩽\left({\alpha }_1{x}_1+{\alpha }_2{x}_2-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right).$$

Lemma 6

([6], Theorem 1). Let $n⩾2$, $x_1,\dots ,x_n⩾0$ and ${\alpha }_1,\dots ,{\alpha }_n>0$. Suppose that ${\alpha }_1+\dots +{\alpha }_n=1$. Then we have

$$\sum _{1⩽i⩽n}{\alpha }_i{\left(x_i^{{\displaystyle \frac{1}{2}}}-\sum _{1⩽j⩽n}{\alpha }_j{x}_j^{{\displaystyle \frac{1}{2}}}\right)}^2⩽\sum _{1⩽i⩽n}{\alpha }_i{x}_i-\prod _{1⩽i⩽n}{x}_i^{{\alpha }_i}.$$

Moreover, for $n=2$ this becomes

$${\alpha }_1{\left(x_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2+{\alpha }_2{\left(x_2^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2⩽\left({\alpha }_1{x}_1+{\alpha }_2{x}_2-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right).$$

Lemma 7

([8], Theorem 2.2). Let $n⩾2$, $x_1,\dots ,x_n⩾0$ and ${\alpha }_1,\dots ,{\alpha }_n>0$. Suppose that ${\alpha }_1+\dots +{\alpha }_n=1$. Then we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{1-min({\alpha }_1,\dots ,{\alpha }_n)}}\sum _{1⩽i⩽n}{\alpha }_i{\left(x_i^{{\displaystyle \frac{1}{2}}}-\sum _{1⩽j⩽n}{\alpha }_j{x}_j^{{\displaystyle \frac{1}{2}}}\right)}^2\hfill \\ \hfill ⩽& \sum _{1⩽i⩽n}{\alpha }_i{x}_i-\prod _{1⩽i⩽n}{x}_i^{{\alpha }_i}\hfill \\ \hfill ⩽& {\displaystyle \frac{1}{min({\alpha }_1,\dots ,{\alpha }_n)}}\sum _{1⩽i⩽n}{\alpha }_i{\left(x_i^{{\displaystyle \frac{1}{2}}}-\sum _{1⩽j⩽n}{\alpha }_j{x}_j^{{\displaystyle \frac{1}{2}}}\right)}^2.\hfill \end{array}$$

Moreover, for $n=2$ this becomes

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{1-min({\alpha }_1,{\alpha }_2)}}\left({\alpha }_1{\left(x_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2+{\alpha }_2{\left(x_2^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2\right)\hfill \\ \hfill ⩽& \left({\alpha }_1{x}_1+{\alpha }_2{x}_2-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right)\hfill \\ \hfill ⩽& {\displaystyle \frac{1}{min({\alpha }_1,{\alpha }_2)}}\left({\alpha }_1{\left(x_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2+{\alpha }_2{\left(x_2^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2\right).\hfill \end{array}$$

Lemma 8

([9], Corollary 2.3). Let $n⩾2$, $x_1,\dots ,x_n⩾0$ and ${\alpha }_1,\dots ,{\alpha }_n>0$. Suppose that ${\alpha }_1+\dots +{\alpha }_n=1$. Then we have

$$\begin{array}{cc}\hfill & nmin({\alpha }_1,\dots ,{\alpha }_n)\left({\displaystyle \frac{1}{n}}\sum _{1⩽i⩽n}{x}_i-\prod _{1⩽i⩽n}{x}_i^{{\displaystyle \frac{1}{n}}}\right)\hfill \\ \hfill ⩽& \sum _{1⩽i⩽n}{\alpha }_i{x}_i-\prod _{1⩽i⩽n}{x}_i^{{\alpha }_i}\hfill \\ \hfill ⩽& nmax({\alpha }_1,\dots ,{\alpha }_n)\left({\displaystyle \frac{1}{n}}\sum _{1⩽i⩽n}{x}_i-\prod _{1⩽i⩽n}{x}_i^{{\displaystyle \frac{1}{n}}}\right).\hfill \end{array}$$

Moreover, for $n=2$ this becomes

$$2min({\alpha }_1,{\alpha }_2)\left({\displaystyle \frac{1}{2}}(x_1+x_2)-x_1^{{\displaystyle \frac{1}{2}}}{x}_2^{{\displaystyle \frac{1}{2}}}\right)⩽\left({\alpha }_1{x}_1+{\alpha }_2{x}_2-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right)⩽2max({\alpha }_1,{\alpha }_2)\left({\displaystyle \frac{1}{2}}(x_1+x_2)-x_1^{{\displaystyle \frac{1}{2}}}{x}_2^{{\displaystyle \frac{1}{2}}}\right).$$

Note that the left–hand side is just [[10], Proposition 5.1].

Lemma 9

([11], Theorem 1). Let $n⩾2$, $x_1,\dots ,x_n⩾0$ and ${\alpha }_1,\dots ,{\alpha }_n>0$. Suppose that ${\alpha }_1+\dots +{\alpha }_n=1$. Then we have

$$exp\left(2-2{\displaystyle \frac{{\sum }_{1⩽i⩽n}{\alpha }_i{x}_i^{\frac{1}{2}}}{{\left({\sum }_{1⩽i⩽n}{\alpha }_i{x}_i\right)}^{\frac{1}{2}}}}\right)\prod _{1⩽i⩽n}{x}_i^{{\alpha }_i}⩽\sum _{1⩽i⩽n}{\alpha }_i{x}_i.$$

Moreover, for $n=2$ this becomes

$$exp\left(2-2{\displaystyle \frac{{\alpha }_1{x}_1^{\frac{1}{2}}+{\alpha }_2{x}_2^{\frac{1}{2}}}{{\left({\alpha }_1{x}_1+{\alpha }_2{x}_2\right)}^{\frac{1}{2}}}}\right)x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\alpha }_1{x}_1+{\alpha }_2{x}_2.$$

Lemma 10

([12], Proposition 2.7). Let $n⩾2$, $0⩽x_1⩽\dots ⩽x_j⩽\dots ⩽x_k⩽\dots ⩽x_n$ and ${\alpha }_1,\dots ,{\alpha }_n>0$. Suppose that ${\alpha }_1+\dots +{\alpha }_n=1$. Then we have

$${\displaystyle \frac{3({\alpha }_1+\dots +{\alpha }_j)({\alpha }_k+\dots +{\alpha }_n){(x_k-x_j)}^2}{(4({\alpha }_1+\dots +{\alpha }_j)+2({\alpha }_k+\dots +{\alpha }_n))x_k+(4({\alpha }_k+\dots +{\alpha }_n)+2({\alpha }_1+\dots +{\alpha }_j))x_j}}⩽\sum _{1⩽i⩽n}{\alpha }_i{x}_i-\prod _{1⩽i⩽n}{x}_i^{{\alpha }_i}.$$

Moreover, for $n=2$, $j=1$ and $k=2$, this becomes

$${\displaystyle \frac{3{\alpha }_1{\alpha }_2{(x_2-x_1)}^2}{(4{\alpha }_1+2{\alpha }_2)x_2+(4{\alpha }_2+2{\alpha }_1)x_1}}⩽\left({\alpha }_1{x}_1+{\alpha }_2{x}_2-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right).$$

3. Proof of Theorem 1

3.1. Proof of (A1) and (A2)

By Lemma 4 we know that for $x_1,x_2⩾0$, ${\alpha }_1,{\alpha }_2>0$ and ${\alpha }_1+{\alpha }_2=1$, we have

$$x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\alpha }_1{x}_1+{\alpha }_2{x}_2-{\displaystyle \frac{1}{2max(x_1,x_2)}}\left({\alpha }_1{\left(x_1-{\alpha }_1{x}_1-{\alpha }_2{x}_2\right)}^2+{\alpha }_2{\left(x_2-{\alpha }_1{x}_1-{\alpha }_2{x}_2\right)}^2\right).$$

(7)

For (A1), put $x_1=a$, $x_2=b$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (7) becomes

$$\begin{array}{cc}\hfill & a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}\hfill \\ \hfill ⩽& {\displaystyle \frac{a}{a+b}}a+{\displaystyle \frac{b}{a+b}}b-{\displaystyle \frac{1}{2max(a,b)}}\left({\displaystyle \frac{a}{a+b}}{\left(a-{\displaystyle \frac{a}{a+b}}a-{\displaystyle \frac{b}{a+b}}b\right)}^2+{\displaystyle \frac{b}{a+b}}{\left(b-{\displaystyle \frac{a}{a+b}}a-{\displaystyle \frac{b}{a+b}}b\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{1}{2max(a,b)}}\left({\displaystyle \frac{a}{a+b}}{\left({\displaystyle \frac{a(a+b)-a^2-b^2}{a+b}}\right)}^2+{\displaystyle \frac{b}{a+b}}{\left({\displaystyle \frac{b(a+b)-a^2-b^2}{a+b}}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{1}{2max(a,b)}}\left({\displaystyle \frac{a}{a+b}}{\left({\displaystyle \frac{ab-b^2}{a+b}}\right)}^2+{\displaystyle \frac{b}{a+b}}{\left({\displaystyle \frac{ab-a^2}{a+b}}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{1}{2max(a,b)}}\left({\displaystyle \frac{a{(ab-b^2)}^2+b{(ab-a^2)}^2}{{(a+b)}^3}}\right)\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{1}{2max(a,b)}}\left({\displaystyle \frac{ab{(a-b)}^2}{{(a+b)}^2}}\right).\hfill \end{array}$$

(8)

Let $a=\phi \left(n\right)$ and $b=\psi \left(n\right)$. By (2) we have $a⩽b$, hence $max(a,b)=b$. Putting this into (8), we have

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{1}{2max(a,b)}}\left({\displaystyle \frac{ab{(a-b)}^2}{{(a+b)}^2}}\right)\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{1}{2b}}\left({\displaystyle \frac{ab{(a-b)}^2}{{(a+b)}^2}}\right)\hfill \\ \hfill =& {\displaystyle \frac{(a^2+b^2)(a+b)-\frac{1}{2}a{(a-b)}^2}{{(a+b)}^2}}\hfill \\ \hfill =& {\displaystyle \frac{a^3+4a^{2}b+a{b}^2+2b^3}{2{(a+b)}^2}},\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill a^a{b}^b⩽& {\left({\displaystyle \frac{a^3+4a^{2}b+a{b}^2+2b^3}{2{(a+b)}^2}}\right)}^{a+b}.\hfill \end{array}$$

(10)

Now (A1) is proved. For (A2), put $x_1=b=\psi \left(n\right)$, $x_2=a=\phi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (7) becomes

$$\begin{array}{cc}\hfill & b^{{\displaystyle \frac{a}{a+b}}}{a}^{{\displaystyle \frac{b}{a+b}}}\hfill \\ \hfill ⩽& {\displaystyle \frac{a}{a+b}}b+{\displaystyle \frac{b}{a+b}}a-{\displaystyle \frac{1}{2b}}\left({\displaystyle \frac{a}{a+b}}{\left(b-{\displaystyle \frac{a}{a+b}}b-{\displaystyle \frac{b}{a+b}}a\right)}^2+{\displaystyle \frac{b}{a+b}}{\left(a-{\displaystyle \frac{a}{a+b}}b-{\displaystyle \frac{b}{a+b}}a\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{2ab}{a+b}}-{\displaystyle \frac{1}{2b}}\left({\displaystyle \frac{a}{a+b}}{\left({\displaystyle \frac{b(a+b)-2ab}{a+b}}\right)}^2+{\displaystyle \frac{b}{a+b}}{\left({\displaystyle \frac{a(a+b)-2ab}{a+b}}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{2ab}{a+b}}-{\displaystyle \frac{1}{2b}}\left({\displaystyle \frac{a}{a+b}}{\left({\displaystyle \frac{b^2-ab}{a+b}}\right)}^2+{\displaystyle \frac{b}{a+b}}{\left({\displaystyle \frac{a^2-ab}{a+b}}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{2ab}{a+b}}-{\displaystyle \frac{1}{2b}}\left({\displaystyle \frac{ab{(a-b)}^2}{{(a+b)}^2}}\right)\hfill \\ \hfill =& {\displaystyle \frac{2ab(a+b)-\frac{1}{2}a{(a-b)}^2}{{(a+b)}^2}}\hfill \\ \hfill =& {\displaystyle \frac{-a^3+6a^{2}b+3a{b}^2}{2{(a+b)}^2}},\hfill \end{array}$$

(11)

$$b^a{a}^b⩽{\left({\displaystyle \frac{-a^3+6a^{2}b+3a{b}^2}{2{(a+b)}^2}}\right)}^{a+b}.$$

(12)

Now (A2) is proved.

3.2. Proof of (B1) and (B2)

By Lemma 5 we know that for $x_1,x_2⩾0$, ${\alpha }_1,{\alpha }_2>0$ and ${\alpha }_1+{\alpha }_2=1$, we have

$$x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\alpha }_1{x}_1+{\alpha }_2{x}_2-{\displaystyle \frac{1}{2max(x_1,x_2)}}\left({\alpha }_1{\left(x_1-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right)}^2+{\alpha }_2{\left(x_2-x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\right)}^2\right).$$

(13)

For (B1), put $x_1=a=\phi \left(n\right)$, $x_2=b=\psi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (13) becomes

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}a+{\displaystyle \frac{b}{a+b}}b-{\displaystyle \frac{1}{2b}}\left({\displaystyle \frac{a}{a+b}}{\left(a-a^{{\displaystyle \frac{a}{a+b}}}-b^{{\displaystyle \frac{b}{a+b}}}\right)}^2+{\displaystyle \frac{b}{a+b}}{\left(b-a^{{\displaystyle \frac{a}{a+b}}}-b^{{\displaystyle \frac{b}{a+b}}}\right)}^2\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{a{\left(a-a^{\frac{a}{a+b}}-b^{\frac{b}{a+b}}\right)}^2+b{\left(b-a^{\frac{a}{a+b}}-b^{\frac{b}{a+b}}\right)}^2}{2b(a+b)}},\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill a^a{b}^b⩽& {\left({\displaystyle \frac{2a^{2}b+2b^3-a{\left(a-a^{\frac{a}{a+b}}-b^{\frac{b}{a+b}}\right)}^2-b{\left(b-a^{\frac{a}{a+b}}-b^{\frac{b}{a+b}}\right)}^2}{2b(a+b)}}\right)}^{a+b}.\hfill \end{array}$$

(15)

Now (B1) is proved. For (B2), put $x_1=b=\psi \left(n\right)$, $x_2=a=\phi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (13) becomes

$$\begin{array}{cc}\hfill b^{{\displaystyle \frac{a}{a+b}}}{a}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}b+{\displaystyle \frac{b}{a+b}}a-{\displaystyle \frac{1}{2b}}\left({\displaystyle \frac{a}{a+b}}{\left(b-b^{{\displaystyle \frac{a}{a+b}}}-a^{{\displaystyle \frac{b}{a+b}}}\right)}^2+{\displaystyle \frac{b}{a+b}}{\left(a-b^{{\displaystyle \frac{a}{a+b}}}-a^{{\displaystyle \frac{b}{a+b}}}\right)}^2\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{2ab}{a+b}}-{\displaystyle \frac{a{\left(b-b^{\frac{a}{a+b}}-a^{\frac{b}{a+b}}\right)}^2+b{\left(a-b^{\frac{a}{a+b}}-a^{\frac{b}{a+b}}\right)}^2}{2b(a+b)}},\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill b^a{a}^b⩽& {\left({\displaystyle \frac{4a{b}^2-a{\left(b-b^{\frac{a}{a+b}}-a^{\frac{b}{a+b}}\right)}^2-b{\left(a-b^{\frac{a}{a+b}}-a^{\frac{b}{a+b}}\right)}^2}{2b(a+b)}}\right)}^{a+b}.\hfill \end{array}$$

(17)

Now (B2) is proved.

3.3. Proof of (C1) and (C2)

By Lemma 6 we know that for $x_1,x_2⩾0$, ${\alpha }_1,{\alpha }_2>0$ and ${\alpha }_1+{\alpha }_2=1$, we have

$$x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\alpha }_1{x}_1+{\alpha }_2{x}_2-\left({\alpha }_1{\left(x_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2+{\alpha }_2{\left(x_2^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2\right).$$

(18)

For (C1), put $x_1=a=\phi \left(n\right)$, $x_2=b=\psi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (18) becomes

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}a+{\displaystyle \frac{b}{a+b}}b-\left({\displaystyle \frac{a}{a+b}}{\left(a^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{a}{a+b}}{a}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{b}{a+b}}{b}^{{\displaystyle \frac{1}{2}}}\right)}^2\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{b}{a+b}}{\left(b^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{a}{a+b}}{a}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{b}{a+b}}{b}^{{\displaystyle \frac{1}{2}}}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{ab\left(a+b-2a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)}{{(a+b)}^2}}\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill =& {\left({\displaystyle \frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}}{a+b}}\right)}^2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill a^a{b}^b⩽& {\left({\displaystyle \frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}}{a+b}}\right)}^{2(a+b)}.\hfill \end{array}$$

(20)

Now (C1) is proved. For (C2), put $x_1=b=\psi \left(n\right)$, $x_2=a=\phi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (18) becomes

$$\begin{array}{cc}\hfill b^{{\displaystyle \frac{a}{a+b}}}{a}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}b+{\displaystyle \frac{b}{a+b}}a-\left({\displaystyle \frac{a}{a+b}}{\left(b^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{a}{a+b}}{b}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{b}{a+b}}{a}^{{\displaystyle \frac{1}{2}}}\right)}^2\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{b}{a+b}}{\left(a^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{a}{a+b}}{b}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{b}{a+b}}{a}^{{\displaystyle \frac{1}{2}}}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{2ab}{a+b}}-{\displaystyle \frac{ab\left(a+b-2a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)}{{(a+b)}^2}}\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill =& {\left({\displaystyle \frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}{a+b}}\right)}^2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill b^a{a}^b⩽& {\left({\displaystyle \frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}{a+b}}\right)}^{2(a+b)}.\hfill \end{array}$$

(22)

Now (C2) is proved.

3.4. Proof of (D1) and (D2)

By Lemma 7 we know that for $x_1,x_2⩾0$, ${\alpha }_1,{\alpha }_2>0$ and ${\alpha }_1+{\alpha }_2=1$, we have

$$x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\alpha }_1{x}_1+{\alpha }_2{x}_2-{\displaystyle \frac{1}{1-min({\alpha }_1,{\alpha }_2)}}\left({\alpha }_1{\left(x_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2+{\alpha }_2{\left(x_2^{{\displaystyle \frac{1}{2}}}-{\alpha }_1{x}_1^{{\displaystyle \frac{1}{2}}}-{\alpha }_2{x}_2^{{\displaystyle \frac{1}{2}}}\right)}^2\right).$$

(23)

For (D1), we put $x_1=a=\phi \left(n\right)$, $x_2=b=\psi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$. Since $a+b>0$ and $a⩽b$, we have

$${\alpha }_1={\displaystyle \frac{a}{a+b}}⩽{\displaystyle \frac{b}{a+b}}={\alpha }_2,$$

hence $1-min({\alpha }_1,{\alpha }_2)=max({\alpha }_1,{\alpha }_2)={\alpha }_2={\displaystyle \frac{b}{a+b}}$. Then (23) becomes

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}a+{\displaystyle \frac{b}{a+b}}b-{\displaystyle \frac{1}{\frac{b}{a+b}}}\left({\displaystyle \frac{a}{a+b}}{\left(a^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{a}{a+b}}{a}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{b}{a+b}}{b}^{{\displaystyle \frac{1}{2}}}\right)}^2\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{b}{a+b}}{\left(b^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{a}{a+b}}{a}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{b}{a+b}}{b}^{{\displaystyle \frac{1}{2}}}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{a+b}{b}}·{\displaystyle \frac{ab\left(a+b-2a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)}{{(a+b)}^2}}\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{a\left(a+b-2a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)}{a+b}}\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{2a^{\frac{3}{2}}{b}^{\frac{1}{2}}-ab+b^2}{a+b}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill a^a{b}^b⩽& {\left({\displaystyle \frac{2a^{\frac{3}{2}}{b}^{\frac{1}{2}}-ab+b^2}{a+b}}\right)}^{a+b}.\hfill \end{array}$$

(25)

Now (D1) is proved. For (D2), put $x_1=b=\psi \left(n\right)$, $x_2=a=\phi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (23) becomes

$$\begin{array}{cc}\hfill b^{{\displaystyle \frac{a}{a+b}}}{a}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}b+{\displaystyle \frac{b}{a+b}}a-{\displaystyle \frac{1}{\frac{b}{a+b}}}\left({\displaystyle \frac{a}{a+b}}{\left(b^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{a}{a+b}}{b}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{b}{a+b}}{a}^{{\displaystyle \frac{1}{2}}}\right)}^2\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{b}{a+b}}{\left(a^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{a}{a+b}}{b}^{{\displaystyle \frac{1}{2}}}-{\displaystyle \frac{b}{a+b}}{a}^{{\displaystyle \frac{1}{2}}}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{2ab}{a+b}}-{\displaystyle \frac{a\left(a+b-2a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)}{a+b}}\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{2a^{\frac{3}{2}}{b}^{\frac{1}{2}}+ab-a^2}{a+b}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill b^a{a}^b⩽& {\left({\displaystyle \frac{2a^{\frac{3}{2}}{b}^{\frac{1}{2}}+ab-a^2}{a+b}}\right)}^{a+b}.\hfill \end{array}$$

(27)

Now (D2) is proved.

We note that Lemma 8 and Lemma 7 actually yield the same result here. By Lemma 8 we know that for $x_1,x_2⩾0$, ${\alpha }_1,{\alpha }_2>0$ and ${\alpha }_1+{\alpha }_2=1$, we have

$$x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\alpha }_1{x}_1+{\alpha }_2{x}_2-\left(2min({\alpha }_1,{\alpha }_2)\left({\displaystyle \frac{1}{2}}(x_1+x_2)-x_1^{{\displaystyle \frac{1}{2}}}{x}_2^{{\displaystyle \frac{1}{2}}}\right)\right).$$

(28)

For (D1), we put $x_1=a=\phi \left(n\right)$, $x_2=b=\psi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$. Again, we have $min({\alpha }_1,{\alpha }_2)={\alpha }_1={\displaystyle \frac{a}{a+b}}$. Then (28) becomes

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}a+{\displaystyle \frac{b}{a+b}}b-2{\displaystyle \frac{a}{a+b}}\left({\displaystyle \frac{1}{2}}a+{\displaystyle \frac{1}{2}}b-a^{{\displaystyle \frac{1}{2}}}{b}^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2-2a\left(\frac{1}{2}a+\frac{1}{2}b-a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)}{a+b}}\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{2a^{\frac{3}{2}}{b}^{\frac{1}{2}}-ab+b^2}{a+b}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill a^a{b}^b⩽& {\left({\displaystyle \frac{2a^{\frac{3}{2}}{b}^{\frac{1}{2}}-ab+b^2}{a+b}}\right)}^{a+b}.\hfill \end{array}$$

(30)

Now (D1) is proved. For (D2), put $x_1=b=\psi \left(n\right)$, $x_2=a=\phi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (28) becomes

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}b+{\displaystyle \frac{b}{a+b}}a-2{\displaystyle \frac{a}{a+b}}\left({\displaystyle \frac{1}{2}}b+{\displaystyle \frac{1}{2}}a-b^{{\displaystyle \frac{1}{2}}}{a}^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill =& {\displaystyle \frac{2ab-2a\left(\frac{1}{2}a+\frac{1}{2}b-a^{\frac{1}{2}}{b}^{\frac{1}{2}}\right)}{a+b}}\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{2a^{\frac{3}{2}}{b}^{\frac{1}{2}}+ab-a^2}{a+b}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill b^a{a}^b⩽& {\left({\displaystyle \frac{2a^{\frac{3}{2}}{b}^{\frac{1}{2}}+ab-a^2}{a+b}}\right)}^{a+b}.\hfill \end{array}$$

(32)

Now (D2) is proved.

3.5. Proof of (E1) and (E2)

By Lemma 9 we know that for $x_1,x_2⩾0$, ${\alpha }_1,{\alpha }_2>0$ and ${\alpha }_1+{\alpha }_2=1$, we have

$$x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\displaystyle \frac{{\alpha }_1{x}_1+{\alpha }_2{x}_2}{exp\left(2-2\frac{{\alpha }_1{x}_1^{\frac{1}{2}}+{\alpha }_2{x}_2^{\frac{1}{2}}}{{\left({\alpha }_1{x}_1+{\alpha }_2{x}_2\right)}^{\frac{1}{2}}}\right)}}.$$

(33)

For (E1), put $x_1=a=\phi \left(n\right)$, $x_2=b=\psi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (33) becomes

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{\frac{a}{a+b}a+\frac{b}{a+b}b}{exp\left(2-2\frac{\frac{a}{a+b}{a}^{\frac{1}{2}}+\frac{b}{a+b}{b}^{\frac{1}{2}}}{{\left(\frac{a}{a+b}a+\frac{b}{a+b}b\right)}^{\frac{1}{2}}}\right)}}\hfill \end{array}$$

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{a^2+b^2}{(a+b)exp\left(2-2\frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}}{{\left(\left(a^2+b^2\right)(a+b)\right)}^{\frac{1}{2}}}\right)}},\hfill \end{array}$$

(34)

$$\begin{array}{cc}\hfill a^a{b}^b⩽& {\left({\displaystyle \frac{a^2+b^2}{(a+b)exp\left(2-2\frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}}{{\left(\left(a^2+b^2\right)(a+b)\right)}^{\frac{1}{2}}}\right)}}\right)}^{a+b}.\hfill \end{array}$$

(35)

Now (E1) is proved. For (E2), put $x_1=b=\psi \left(n\right)$, $x_2=a=\phi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$, (33) becomes

$$\begin{array}{cc}\hfill b^{{\displaystyle \frac{a}{a+b}}}{a}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{\frac{a}{a+b}b+\frac{b}{a+b}a}{exp\left(2-2\frac{\frac{a}{a+b}{b}^{\frac{1}{2}}+\frac{b}{a+b}{a}^{\frac{1}{2}}}{{\left(\frac{a}{a+b}b+\frac{b}{a+b}a\right)}^{\frac{1}{2}}}\right)}}\hfill \end{array}$$

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{2ab}{(a+b)exp\left(2-2\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}{{\left(2ab(a+b)\right)}^{\frac{1}{2}}}\right)}},\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill b^a{a}^b⩽& {\left({\displaystyle \frac{2ab}{(a+b)exp\left(2-2\frac{a^{\frac{1}{2}}{b}^{\frac{1}{2}}\left(a^{\frac{1}{2}}+b^{\frac{1}{2}}\right)}{{\left(2ab(a+b)\right)}^{\frac{1}{2}}}\right)}}\right)}^{a+b}.\hfill \end{array}$$

(37)

Now (E2) is proved.

3.6. Proof of (F1) and (F2)

By Lemma 10 we know that for $0⩽x_1⩽x_2$, ${\alpha }_1,{\alpha }_2>0$ and ${\alpha }_1+{\alpha }_2=1$, we have

$$x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}⩽{\alpha }_1{x}_1+{\alpha }_2{x}_2-{\displaystyle \frac{3{\alpha }_1{\alpha }_2{(x_2-x_1)}^2}{(4{\alpha }_1+2{\alpha }_2)x_2+(4{\alpha }_2+2{\alpha }_1)x_1}}.$$

(38)

For (F1), we put $x_1=a=\phi \left(n\right)$, $x_2=b=\psi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{a}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{b}{a+b}}$. Since $a⩽b$, we can write (38) as

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{a}{a+b}}}{b}^{{\displaystyle \frac{b}{a+b}}}⩽& {\displaystyle \frac{a}{a+b}}a+{\displaystyle \frac{b}{a+b}}b-{\displaystyle \frac{3\frac{a}{a+b}·\frac{b}{a+b}{(b-a)}^2}{(4\frac{a}{a+b}+2\frac{b}{a+b})b+(4\frac{b}{a+b}+2\frac{a}{a+b})a}}\hfill \\ \hfill =& {\displaystyle \frac{a^2+b^2}{a+b}}-{\displaystyle \frac{\frac{3ab{(b-a)}^2}{{(a+b)}^2}}{\frac{2a^2+8ab+2b^2}{a+b}}}\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{a^2+b^2-\frac{3ab{(b-a)}^2}{2a^2+8ab+2b^2}}{a+b}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill a^a{b}^b⩽& {\left({\displaystyle \frac{a^2+b^2-\frac{3ab{(b-a)}^2}{2a^2+8ab+2b^2}}{a+b}}\right)}^{a+b}.\hfill \end{array}$$

(40)

Now (F1) is proved. For (F2), we put $x_1=a=\phi \left(n\right)$, $x_2=b=\psi \left(n\right)$, ${\alpha }_1={\displaystyle \frac{b}{a+b}}$ and ${\alpha }_2={\displaystyle \frac{a}{a+b}}$. Since $a⩽b$, (38) becomes

$$\begin{array}{cc}\hfill a^{{\displaystyle \frac{b}{a+b}}}{b}^{{\displaystyle \frac{a}{a+b}}}⩽& {\displaystyle \frac{b}{a+b}}a+{\displaystyle \frac{a}{a+b}}b-{\displaystyle \frac{3\frac{b}{a+b}·\frac{a}{a+b}{(b-a)}^2}{(4\frac{b}{a+b}+2\frac{a}{a+b})b+(4\frac{a}{a+b}+2\frac{b}{a+b})a}}\hfill \\ \hfill =& {\displaystyle \frac{2ab}{a+b}}-{\displaystyle \frac{\frac{3ab{(b-a)}^2}{{(a+b)}^2}}{\frac{4a^2+4ab+4b^2}{a+b}}}\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill =& {\displaystyle \frac{2ab-\frac{3ab{(b-a)}^2}{4\left(a^2+ab+b^2\right)}}{a+b}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill a^b{b}^a⩽& {\left({\displaystyle \frac{2ab-\frac{3ab{(b-a)}^2}{4\left(a^2+ab+b^2\right)}}{a+b}}\right)}^{a+b}.\hfill \end{array}$$

(42)

Now (F2) is proved.