I have this to solve :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c$ then : $$\tan(a)^b+\tan(b)^c+\tan(c)^a\geq \tan(a)^c+\tan(c)^b+\tan(b)^a$$
I try to study the behavior of (with the condition on $a,b$): $$\tan(a)^b-\tan(b)^a$$
And for $a\geq b$ and $a+b<1$ I think we have :
$$\tan(a)^b-\tan(b)^a\geq a^b-b^a$$
Unfortunatly we can't apply this directly because there is the cyclicity (order on $a$ and $b$)
On the other hand I suggest that we have :
For $a\geq b >0$ $\quad$ and $\frac{1}{3}\geq b>0$ and $a+b<1$ we get :
$$2(\tan(a)^{\frac{1}{3}}-tan(b)^{\frac{1}{3}})\geq \tan(a)^b-\tan(b)^a$$
My conclusion is : this is this kind of inequality of $2$ variables wich makes the problem a little bit easier .
My question :Have you nice proof of this fact ?
Thanks a lot for sharing your time and knowledge .
$a=b=c$ obviously works. From this position, you are allowed to increase $a$ and decrease $b$ and $c$. Suppose $a>\frac{1}{3}$ with $b=c < \frac{1}{3}$. Then the inequality is satisfied (you can try this for yourself).
Now let's consider the case $a\geq b \geq c$. We start by the $a>b=c$ inequality and reduce $c$, fixing $b$ and increasing $a$. (This will cover every single case of $a \geq b \geq c$).
We may rewrite by removing $a$ for $1-b-c$:
$f(c) = \tan(1-b-c)^b +\tan(b)^c +\tan(c)^{1-b-c} - \tan(1-b-c)^c - \tan(c)^{b} -\tan(b)^{1-b-c}$.
Now we compute the derivative with respect to $c$, keeping $b$ as a constant.
We note that $f'(c) \leq 0$ here thus the inequality still holds when $c$ is reduced. Hence we have our result.
Below is the proof that $f'(c) \leq 0$.
$f'(c)$ has 8 terms, labelled $t_1, ..., t_8$.
$t_1+t_6 \leq 0$
$t_2+t_8 \leq 0$.
$t_3+t_5 \leq 0$ (here we use that $x\tan^x(k)$ is an increasing function for $\tan(k)<1$)
$t_4+t_7 \leq 0$