Prove $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$

Problem: Let $x, y > 0$. Prove that $$f(x, y) = 2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0.$$

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My attempt:

For $a > 0$, we need to prove that $$f\left(x\right)=(x+a)\ln\frac{a+x}{2}-\frac{\left(a+1\right)\ln a+\left(x+1\right)\ln x}{2} \ge 0.$$

WLOG assume that $a\leq x$

The second derivative can be rewrite as :

$$a(1-x)+x(x+1)\geq 0$$

Wich is obvious with the contraint above .

We deduce that the derivative admits a single zero and is increasing .

Edit to clarify the problem :

It's a straightforward consequence of my answer here show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$ in the final version .

Well after MartinR comment :I need to show it to conclude my answer .

Other idea Case $0<a\leq 1$ and $x\geq a$:

Using simple bound got from https://www.researchgate.net/publication/267163352_Proofs_of_three_open_inequalities_with_power-exponential_functions lemma 7.2 where $b=0$ and $d=\frac{x+a}{2}$ we have the new function :

$$h(x)=2\left(\frac{\left(x+a\right)}{2}-1\right)-0.5\left(\left(x+1\right)\ln\left(x\right)+\left(a+1\right)\ln\left(a\right)\right)$$

It's NOT true that $h(x)\geq 0$ in this case . See the remarks below

Some remarks :

The derivative does not depends from $a$ .

the abscissa of the minimum of $f(x)$ denoted by $x_{min}$ is less or equal than the $x_0$ where $h(x_0)=0$ for $0<a\leq 1$ and $x\geq a$

Important remark :

It seems we have for $0<a\leq 1$ and $x\geq a$:

$$x_{min}\leq 2-a+0.5\left(a-1\right)^{2}\leq x_{0}\quad (I)$$

It reduces the problem to a single variable inequality in $a$ .

Some Conjectures :

Firstly It seems we have $a\in (0,1]$:

$$0\leq f'\left(2-a+0.5\left(a-1\right)^{2}\right)$$

Secondly It seems we have $a\in (0,1]$ :

$$0\leq h\left(2-a+0.5\left(a-1\right)^{2}\right)$$

Question :

How to prove it ?

Thanks.

Problem: Let $x, y > 0$. Prove that $$f(x, y) = 2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0.$$

Proof:

Note that $f(1, 1) = 0$. We need to prove that $x = y = 1$ is the global minimizer of $f(x, y)$ on $x,y > 0$.

We have \begin{align*} f(x, y) &\ge 2x \ln \frac{x}{2} + 2y\ln \frac{y}{2} - (x + 1)\ln x - (y + 1)\ln y\\ &= (x - 1)\ln x - 2x\ln 2 + (y - 1)\ln y - 2y\ln 2. \tag{1} \end{align*}

Fact 1: $(u - 1)\ln u - 2u\ln 2 > -3$ for all $u > 0$.
(The proof is given at the end.)

Fact 2: $(u - 1)\ln u - 2u\ln 2 > 4$ for all $u\in (0, 1/64]\cup [16, \infty)$.
(The proof is given at the end.)

By (1) and Facts 1-2, if $x < 1/64$ or $x > 16$, then $f(x, y) > -3 + 4 = 1$, and if $y < 1/64$ or $y > 16$, then $f(x, y) > -3 + 4 = 1$.

Since $f(1, 1) = 0$, the minimum of $f(x, y)$ on $x, y > 0$ occurs on the region $1/64 \le x, y \le 16$.

The minimum of $f(x, y)$ on the region $1/64 \le x, y \le 16$ may occur in the interior of the region (stationary points), or it may occur on the boundary of the region.

First, we can prove that $x = y = 1$ is the only stationary point of $f(x, y)$ on $x, y > 0$:

Fact 3: If $x, y > 0$ satisfies $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$, then $x = y = 1$.
(The proof is given at the end.)

Second, at the boundaries of the region $1/64 \le x, y \le 16$, it is easy to verify that $f(x, y) > 1$ using (1) and Facts 1-2.

As a result, $f(x, y) \ge 0$ for all $1/64 \le x, y \le 16$.

Thus, $f(x, y) \ge 0$ for all $x, y > 0$.

We are done.

Proof of Fact 1:

Let $h(u) = (u - 1)\ln u - 2u\ln 2 + 3$. Since $(u - 1)\ln u \ge 0$ for all $u > 0$, we have $h(u) > 0$ for all $0 < u < \frac{3}{2\ln 2}$. If $u \ge \frac{3}{2\ln 2}$, using $\ln 2 < \frac{4}{5}$ and $\ln z \ge \frac{2(z - 1)}{z + 1}$ for all $z \ge 1 $, we have $$h(u) \ge (u - 1)\cdot \frac{2(u - 1)}{u + 1} - 2u\cdot \frac{4}{5} + 3 = \frac{2u^2 - 13u + 25}{5u + 5} > 0.$$

We are done.

Proof of Fact 2:

Let $H(u) = (u - 1)\ln u - 2u\ln 2 - 4$. We have $H'(u) = \ln u + 1 - \frac{1}{u} - 2\ln 2$.

If $0 < u \le \frac{1}{64}$, we have $H'(u) < 0$ and thus $H(u) \ge H(1/64) = \frac{47}{8}\ln 2 - 4 > 0$.

If $u \ge 16$, we have $H'(u) > 0$ and thus $H(u) \ge H(16) = 28\ln 2 - 4 > 0$.

We are done.

Proof of Fact 3:

We have \begin{align*} \frac{\partial f}{\partial x} &= 2\ln\frac{x + y}{2} + 1 - \ln x - \frac{1}{x}, \tag{2}\\ \frac{\partial f}{\partial y} &= 2\ln\frac{x + y}{2} + 1 - \ln y - \frac{1}{y}. \tag{3} \end{align*}

(2) gives $$y = -x + 2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}}. \tag{4}$$

[(2) - (3)] gives $$\ln x + \frac{1}{x} = \ln y + \frac{1}{y}. \tag{5}$$

From (4) and (5), we have $$\ln x + \frac{1}{x} - \ln(-x + 2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}}) - \frac{1}{-x + 2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}}} = 0.$$

From $-x + 2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}} > 0$, we have $\frac{1}{x}\mathrm{e}^{1/x} > \mathrm{e}/4$ which results in $0 < x < x_0$ where $x_0$ is the unique real root of $\frac{1}{x}\mathrm{e}^{1/x} = \mathrm{e}/4$.

Let $$g(x) = \ln x + \frac{1}{x} - \ln(-x + 2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}}) - \frac{1}{-x + 2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}}}.$$ We have $$g'(x) = - \frac{\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}}}{x^2(-x + 2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}})^2}\cdot h(x) - \frac{(x - 1)^2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}}}{x^2(-x + 2\sqrt{\mathrm{e}^{-1}x\mathrm{e}^{1/x}})^2} $$ where \begin{align*} h(x) &= 2\sqrt{x}\,(1 - x)(\mathrm{e}^{-1/2 + 1/(2x)} - 1) + \sqrt{x}\,(1 - \sqrt{x})^2\\ &\quad + 2x\sqrt{x}\left[\mathrm{e}^{1/2-1/(2x)} - 1 - \left(\frac{1}{2} - \frac{1}{2x}\right)\right]. \end{align*}

Since $(1 - x)(\mathrm{e}^{-1/2 + 1/(2x)} - 1) \ge 0$ for all $x > 0$, and $\mathrm{e}^{1/2-1/(2x)} - 1 - \left(\frac{1}{2} - \frac{1}{2x}\right) \ge 0$ for all $x > 0$, we have $h(x) \ge 0$ for all $x > 0$.

Thus, $g'(x) < 0$ for all $x \in (0, 1)\cup (1, x_0)$. Also, $g'(1) = 0$ and $g(1) = 0$. Thus, $g(x) = 0$ has exactly one real root, say $x = 1$.

From (4), we have $y = 1$.

We are done.