Problem with an inequality : $\sqrt{\frac{a^x}{x}} - \frac{a-1}{a+1} + \frac{x-1}{x+1}-1\geq 0 \,?$

Question

Hi I ask for a proof about the inequality :

Let $a,x>0$ . Does one have the following inequalities

Let $a,x>1$ or $a\geq 1\geq x$ or $a\leq 1\leq x$ or $0.1\leq a\leq 1$ and $0<x\leq 1$ :

$$\sqrt{\frac{a^x}{x}} - \frac{a-1}{a+1} + \frac{x-1}{x+1}-1\geq 0 \,?$$

I can show it for $a,x> 1$ using Bernoulli's inequality .Wwe need to show :

$$\frac{1+\left(a-1\right)x^{2}}{x}-\frac{a^{2}-1}{a^{2}+1}+\frac{x^{2}-1}{x^{2}+1}-1\geq 0$$

This last inequality is not hard .

Using RiverLi's answer here (show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$) we need to show in the case $a\geq 1\geq x$ :

$$\frac{1}{x}\frac{1+a+(a-1)x^{2}}{1+a-(a-1)x^{2}}-\frac{\left(a^{2}-1\right)}{a^{2}+1}+\frac{\left(x^{2}-1\right)}{x^{2}+1}-1\geq 0$$

This last inequality is again not hard

Another case with $x\leq a\leq 1$ and $a\geq 0.1$ :

Using the well-know inequality $e^x\geq x+1$ on $(-\infty,\infty)$ it seems we have :

$$\frac{1+\ln a^{x^{2}}}{x}-\frac{\left(a^{2}-1\right)}{a^{2}+1}+\frac{\left(x^{2}-1\right)}{x^{2}+1}-1\geq 0$$

I go a little bit further using Bernoulli's inequality (with a trick ) we have for $1\geq x\geq a \geq 0.2$ :

$$\frac{a}{x}\cdot\frac{1}{\left(1-\left(a-1\right)\left(x^{2}-1\right)\right)}-\frac{\left(a^{2}-1\right)}{\left(a^{2}+1\right)}+\frac{\left(x^{2}-1\right)}{x^{2}+1}-1\geq 0$$

This inequality is smooth .

Last edit :

It seems we have for $2\geq x\geq 1$ and $1\geq a$ :

$$\frac{1}{x}\frac{1+a+(a-1)x^{2}}{1+a-(a-1)x^{2}}-\frac{\left(a^{2}-1\right)}{a^{2}+1}+\frac{\left(x^{2}-1\right)}{x^{2}+1}-1\geq 0$$

How to show the rest or find a counter-example ?

Thanks !

The usual questions: Where does the problem come from? Why do you think that those inequalities hold? Why would that be useful or interesting? – And why is it relevant that $1/a$ is not an integer? — Martin R, Jan 24 '22 at 08:19
Now you have removed the first inequality and added a restriction $a> 0.1$ to the second in inequality. All this looks quite random to me, I still cannot see the motivation and context for the question. — Martin R, Jan 24 '22 at 09:52
@MartinR it's simple I try to show the inequality https://math.stackexchange.com/questions/4278217/show-that-left-fracx-1x-2x-2-rightp-left-fracx-2x-3x-3-righ?noredirect=1&lq=1 . — Miss and Mister cassoulet char, Jan 24 '22 at 09:58

Miss and Mister cassoulet char · Answer 1 · 2022-01-25T16:26:21.143

I post a curious fact about RiverLi's lemma :

Using the fact that with Bernoulli's we have :

$$a^{0.5x^{2}}\ge a^{0.5}\cdot\frac{1}{\left(1-0.5\left(a-1\right)\left(x^{2}-1\right)\right)}$$

And again with Bernoulli's :

$$a^{0.5x^{2}}\ge\frac{\left(1+0.5\left(a-1\right)\left(x^{2}+1\right)\right)}{a^{0.5}}$$

Remarking that $x^2=0.5x^2+0.5x^2$ so multiplying these two inequalities we get the desired result on a certain interval by example $a=6$ and $x\in [2,\infty)$

Alternative sketch of proof for $0<x\leq 1\leq a$

Using derivative we have the inequality :

$$\left(1+\left(a\cdot\frac{1}{1-\left(a-1\right)\left(x-1\right)}-1\right)x\right)\leq a^{x^2}$$

Then we need to show :

$$\frac{\left(1+\left(a\cdot\frac{1}{1-\left(a-1\right)\left(x-1\right)}-1\right)x\right)}{x}-\frac{\left(a^{2}-1\right)}{\left(a^{2}+1\right)}-1+\frac{\left(x^{2}-1\right)}{x^{2}+1}\geq 0$$

Wich is not really hard .

Problem with an inequality : $\sqrt{\frac{a^x}{x}} - \frac{a-1}{a+1} + \frac{x-1}{x+1}-1\geq 0 \,?$

1 Answers1

Linked