Challenging inequality with three variables

Question

I'm interested by the following problem :

Let $a,b,c$ be real positive numbers such that $abc=1$ with and $\beta>1$ and $0<\alpha<1$ then : $$\Big(\frac{\alpha a}{a^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha b}{b^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha c}{c^{11}+1}\Big)^\frac{1}{\beta}\leq 3\Big(\frac{\alpha}{2}\Big)^\frac{1}{\beta}$$

I claim that the maximum is reached for the triplet $(1;1;1)$ But I can't prove it ..

Any helps or hints would be appreciated .

Edit :

We start with the case $a\leq 1$ , $b\leq 1$ , $c\geq 1$ so we have : $$\Big(\frac{\alpha a}{a^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha b}{b^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha c}{c^{11}+1}\Big)^\frac{1}{\beta}$$ Or with $a\geq 1$, $b\geq 1$ , $c\leq 1$ : $$\Big(\frac{\alpha a^{10}}{a^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha b^{10}}{b^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha c^{10}}{c^{11}+1}\Big)^\frac{1}{\beta}$$ We have the following lemma :

Let $a,b$ be real positive numbers with $a\geq 1$, $b\geq 1$ then we have : $$\Big(\frac{\alpha a^{10}}{a^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha b^{10}}{b^{11}+1}\Big)^\frac{1}{\beta}\leq2\Big(\frac{\frac{\alpha a^{10}}{a^{11}+1}+\frac{\alpha b^{10}}{b^{11}+1}}{2}\Big)^\frac{1}{\beta}\leq 2\Big(\frac{a+b}{2ab}\frac{\alpha(\frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}\Big)^\frac{1}{\beta}$$

Proof :

It's just the inequality of Jensen apply to $f(x)$ wich is concave for $x\geq 1$ :

$f(x)=\Big(\frac{\alpha x^{11}}{x^{11}+1}\Big)$

The variable are :

$x_1=a$ and $x_2=b$

With coefficient :

$\alpha_1=\frac{1}{a}\frac{ab}{a+b}$

And

$\alpha_2=\frac{1}{b}\frac{ab}{a+b}$

We have this other lemma :

$$\Big(\frac{\alpha c^{10}}{c^{11}+1}\Big)^\frac{1}{\beta}=\Big(\frac{\alpha ab}{(ab)^{11}+1}\Big)^\frac{1}{\beta}\leq \Big(\frac{\alpha (\frac{2ab}{a+b})^{2}}{(\frac{2ab}{a+b})^{22}+1}\Big)^\frac{1}{\beta} $$

Proof :

It's easy to show this because $f(x)=\Big(\frac{\alpha x}{x^{11}+1}\Big)$ is decreasing for $x\geq 1$

It's remains to prove : $$(\frac{2ab}{a+b})^{2}\leq ab $$

Wich is obvious.

So we have :

$$2\Big(\frac{a+b}{2ab}\frac{\alpha( \frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha( \frac{2ab}{a+b})^{2}}{(\frac{2ab}{a+b})^{22}+1}\Big)^\frac{1}{\beta}$$

Now we put :

$x=\frac{2ab}{a+b}$

We get for $x\geq 1$:

$$2\Big(\frac{\alpha (x)^{10}}{(x)^{11}+1}\Big)^\frac{1}{\beta}+\Big(\frac{\alpha (x)^{2}}{(x)^{22}+1}\Big)^\frac{1}{\beta}\leq 3\Big(\frac{\alpha}{2}\Big)^\frac{1}{\beta}$$

My questions :

How to get the other cases ?

How to prove this last one variable inequality ?

Have you another way to prove this ?

Answer 1

$\alpha$ can simply be factored out, so we can set it to $1$ and then ignore it.

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At Least Two of $\boldsymbol{a,b,c}$ Must be Equal

To insure that $$ \delta\left[f(a)+f(b)+f(c)\right]=f'(a)\,\delta a+f'(b)\,\delta b+f'(c)\,\delta c=0 $$ for all variations $\delta a,\delta b,\delta c$ so that $$ \delta(abc)=abc\left(\frac{\delta a}{a}+\frac{\delta b}{b}+\frac{\delta c}{c}\right)=0 $$ orthogonality requires a $\lambda$ so that $$ af'(a)=bf'(b)=cf'(c)=\lambda $$ For any $n\in\mathbb{N}$ and $f(x)=\left(\frac{x}{x^{11}+1}\right)^{1/\beta}$, if we look at $xf'(x)$, we see that it is $2$-$1$ everywhere, except at the extreme points. This means that whatever $\lambda$ we have, there are at most two values of $x$ so that $xf'(x)=\lambda$. That is, at least two of $a,b,c$ must be equal.

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Optimal Value of $\boldsymbol{a}$

Without loss of generality, assume that $b=a$ and $c=a^{-2}$. Then we want to maximize $2f(a)+f\!\left(a^{-2}\right)$: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}\left(2f(a)+f\!\left(a^{-2}\right)\right) &=\frac2a\left(af'(a)-a^{-2}f'\!\left(a^{-2}\right)\right)\\ &=0 \end{align} $$ is true when $a=1$. However, if $1\lt\beta\lt1.088$, there are two values of $a$ where the derivative vanishes. Looking at plots, it appears that the critical point at $a=1$ gives the maximum. In fact, if we look at the plot for $\beta=1$, the critical point at $a=1$ gives the maximum:

Greater values of $\beta$ give a smaller maximum for $a\lt1$.

If we accept that $a=b=c=1$ gives the maximum, we get $$ \left(\frac{a}{a^{11}+1}\right)^{1/\beta}+\left(\frac{b}{b^{11}+1}\right)^{1/\beta}+\left(\frac{c}{c^{11}+1}\right)^{1/\beta}\le3\left(\frac12\right)^{1/\beta} $$