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

Question

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 .

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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 .

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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 $a\in (0,1]$ :

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

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Question :

How to prove it ?

Thanks.

Answer 1

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.

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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.

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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.

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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.

Answer 2

Remarks: This is to give an alternative proof of Fact 3 in my answer.
The new proof is related to the following three questions:
Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$,
and
https://artofproblemsolving.com/community/c6h2849102,
and
Lower bound for the square root sum of the roots of $x - \ln x - m$.

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

Fact 3: If $x, y > 0$ satisfies $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$, then $x = y = 1$.

Proof of Fact 3:

We give the following auxiliary result.

Lemma 1: Let $a > 1$. The equation $\frac{1}{z} + \ln z = a$ has two distinct real roots $z_1 < z_2$. Then $2\ln \frac{z_1 + z_2}{2} + 1 > a$. (The proof is given at the end.)

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

From $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$, we have $$2\ln\frac{x + y}{2} + 1 = \ln x + \frac{1}{x} = \ln y + \frac{1}{y}. \tag{1}$$

If $x = y$, from (1), we have $x = y = 1$ (easy).

If $x \ne y$, from $\ln x + \frac{1}{x} = \ln y + \frac{1}{y}$, we have $\ln x + \frac{1}{x} > 1$ (easy). However, by Lemma 1, $2\ln\frac{x + y}{2} + 1 = \ln x + \frac{1}{x}$ is impossible.

As a result, we have $x = y = 1$.

We are done.

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Proof of Lemma 1:

We use the parametrization by wyz@AoPS. Let $t = z_2/z_1 > 1$. We have \begin{align*} \frac{1}{z_1} + \ln z_1 &= a,\\ \frac{1}{tz_1} + \ln(t z_1) &= a \end{align*} which results in $\frac{1}{z_1} = \frac{1}{tz_1} + \ln t$ or $z_1 = \frac{t - 1}{t\ln t}$. Then we have $z_2 = \frac{t-1}{\ln t}$ and $a = \ln \frac{t - 1}{\ln t} + \frac{\ln t}{t - 1}$.

It suffices to prove that, for all $t > 1$, $$2\ln \frac{\frac{t - 1}{t\ln t} + \frac{t-1}{\ln t}}{2} + 1 > \ln \frac{t - 1}{\ln t} + \frac{\ln t}{t - 1}$$ or $$h(t) := 2\ln\frac{1+t}{2t} + \ln \frac{t - 1}{\ln t} + 1 - \frac{\ln t}{t - 1} > 0.$$ We have \begin{align*} h'(t) &= \frac{(t^2 + t)\ln^2 t + (t - 1)^3\ln t - (t + 1)(t - 1)^2}{t(1 + t)(t - 1)^2\ln t}\\[6pt] &> \frac{(t^2 + t)\cdot \frac{2(t - 1)}{t + 1}\cdot\ln t + (t - 1)^3\ln t - (t + 1)(t - 1)^2}{t(1 + t)(t - 1)^2\ln t} \tag{2}\\[6pt] &= \frac{t^2 + 1}{2t(t^2 - 1)\ln t}\left(\ln (t^2) - \frac{2(t^2 - 1)}{t^2 + 1}\right)\\[6pt] &> 0 \tag{3} \end{align*} where we have used $\ln t > \frac{2(t-1)}{t + 1}$ for all $t > 1$ in (2) and (3). Also, we have $\lim_{t\to 1^{+}} h(t) = 0$. Thus, we have $h(t) > 0$ for all $t > 1$.

We are done.

Answer 3

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First conjecture :

Define :

$$F(x)=f'(x)$$

We have $a\in(0,1)$ :

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

Using differentiation we have :

$$(F(2-a+0.5\left(a-1\right)^{2}))'=\frac{\left(a^{5}-10a^{4}+40a^{3}-74a^{2}+63a-20\right)}{\left(a^{2}-4a+5\right)^{2}\left(a^{2}-2a+5\right)}=\frac{\left(a-1\right)^{2}\left(a^{3}-8a^{2}+23a-20\right)}{\left(a^{2}-4a+5\right)^{2}\left(a^{2}-2a+5\right)}<0$$

We are done !

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Second conjecture :

We need to show for $x\in(0,1]$ :

$$p(x)=(x-1)^2-(x+1)\ln(x)-(2-x+0.5(x-1)^2+1)\ln(2-x+0.5(x-1)^2)\geq 0$$

The second derivative is :

$$p''(x)=\frac{-\left(-2x^{2}+8x-6\right)}{\left(x^{2}-4x+5\right)^{2}}+\frac{\left(1-x\right)}{x^{2}}+\frac{2}{\left(x-4\right)x+5}-\ln\left(\left(0.5x-2\right)x+2.5\right)-1$$

Now we use the bound $z\in[1,3]$:

$$k(x)=\frac{0.5\left(z^{2}-1\right)}{z}\geq \ln(z)$$

Remains to show :

$$\frac{-\left(-2x^{2}+8x-6\right)}{\left(x^{2}-4x+5\right)^{2}}+\frac{\left(1-x\right)}{x^{2}}+\frac{2}{\left(x-4\right)x+5}-k\left(\left(0.5x-2\right)x+2.5\right)-1\geq 0$$

Wich is smooth using a computer .

So now as conclusion we have showed that the minimum is greater than zero because the abscissa of the minimum of $f(x)$ is less or equal than the root of $h(x)$ .