Dgdrxrt

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 :
$$Big(fracalpha aa^11+1Big)^frac1beta+Big(fracalpha bb^11+1Big)^frac1beta+Big(fracalpha cc^11+1Big)^frac1betaleq 3Big(fracalpha2Big)^frac1beta$$

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 $aleq 1$ , $bleq 1$ , $cgeq 1$ so we have :
$$Big(fracalpha aa^11+1Big)^frac1beta+Big(fracalpha bb^11+1Big)^frac1beta+Big(fracalpha cc^11+1Big)^frac1beta$$
Or with $ageq 1$, $bgeq 1$ , $cleq 1$ :
$$Big(fracalpha a^10a^11+1Big)^frac1beta+Big(fracalpha b^10b^11+1Big)^frac1beta+Big(fracalpha c^10c^11+1Big)^frac1beta$$
We have the following lemma :

Let $a,b$ be real positive numbers with $ageq 1$, $bgeq 1$ then we have :
$$Big(fracalpha a^10a^11+1Big)^frac1beta+Big(fracalpha b^10b^11+1Big)^frac1betaleq2Big(fracfracalpha a^10a^11+1+fracalpha b^10b^11+12Big)^frac1betaleq 2Big(fraca+b2abfracalpha(frac2aba+b)^11(frac2aba+b)^11+1Big)^frac1beta$$

Proof :

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

$f(x)=Big(fracalpha x^11x^11+1Big)$

The variable are :

$x_1=a$ and $x_2=b$

With coefficient :

$alpha_1=frac1afracaba+b$

And

$alpha_2=frac1bfracaba+b$

We have this other lemma :

$$Big(fracalpha c^10c^11+1Big)^frac1beta=Big(fracalpha ab(ab)^11+1Big)^frac1betaleq Big(fracalpha (frac2aba+b)^2(frac2aba+b)^22+1Big)^frac1beta $$

Proof :

It's easy to show this because $f(x)=Big(fracalpha xx^11+1Big)$ is decreasing for $xgeq 1$

It's remains to prove :
$$(frac2aba+b)^2leq ab $$

Wich is obvious.

So we have :

$$2Big(fraca+b2abfracalpha( frac2aba+b)^11(frac2aba+b)^11+1Big)^frac1beta+Big(fracalpha( frac2aba+b)^2(frac2aba+b)^22+1Big)^frac1beta$$

Now we put :

$x=frac2aba+b$

We get for $xgeq 1$:

$$2Big(fracalpha (x)^10(x)^11+1Big)^frac1beta+Big(fracalpha (x)^2(x)^22+1Big)^frac1betaleq 3Big(fracalpha2Big)^frac1beta$$

My questions :

How to get the other cases ?

How to prove this last one variable inequality ?

Have you another way to prove this ?

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

At Least Two of $boldsymbola,b,c$ Must be Equal

To insure that
$$
deltaleft[f(a)+f(b)+f(c)right]=f'(a),delta a+f'(b),delta b+f'(c),delta c=0
$$
for all variations $delta a,delta b,delta c$ so that
$$
delta(abc)=abcleft(fracdelta aa+fracdelta bb+fracdelta ccright)=0
$$
orthogonality requires a $lambda$ so that
$$
af'(a)=bf'(b)=cf'(c)=lambda
$$
For any $ninmathbbN$ and $f(x)=left(fracxx^11+1right)^1/beta$, 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 $boldsymbola$

Without loss of generality, assume that $b=a$ and $c=a^-2$. Then we want to maximize $2f(a)+f!left(a^-2right)$:
$$
beginalign
fracmathrmdmathrmdaleft(2f(a)+f!left(a^-2right)right)
&=frac2aleft(af'(a)-a^-2f'!left(a^-2right)right)\
&=0
endalign
$$
is true when $a=1$. However, if $1ltbetalt1.088$, 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 $alt1$.

If we accept that $a=b=c=1$ gives the maximum, we get
$$
left(fracaa^11+1right)^1/beta+left(fracbb^11+1right)^1/beta+left(fraccc^11+1right)^1/betale3left(frac12right)^1/beta
$$