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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 ?

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    $\begingroup$ Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. $\endgroup$ – robjohn Mar 25 at 18:13
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    $\begingroup$ Your claim is true. I can provide an answer if more context is provided. $\endgroup$ – robjohn Mar 25 at 22:49
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    $\begingroup$ Are you sure of your statement ?? I don't understand what is the role of $\alpha$, since you can divide the two members of your inequality by $\alpha^{1/\beta}$... $\endgroup$ – TheSilverDoe Mar 29 at 17:46
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    $\begingroup$ You changed the question significantly since my previous comment. There is context, but the answer I wrote no longer applies. If you have a new question, it is best to post another question. $\endgroup$ – robjohn Mar 29 at 17:59
  • $\begingroup$ This music sounds familiar. @FW, it is enough to look at $\alpha=\beta=1$, and then it’s the same as math.stackexchange.com/questions/2602035/… $\endgroup$ – Macavity Apr 1 at 17:55
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$\alpha$ can simply be factored out, so we can set it to $1$ and then ignore it.


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.


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:

enter image description here

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

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

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