One of my old inequality (very sharp)

Question

I'm proud to present one of my old inequality that I can't solve :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\ge b \geq c $ then we have :$$\sqrt{\frac{a}{a^a+b^b}}+\sqrt{\frac{b}{b^b+c^c}}+\sqrt{\frac{c}{c^c+a^a}}\geq \sqrt{\frac{a}{a^a+c^c}}+\sqrt{\frac{c}{c^c+b^b}}+\sqrt{\frac{b}{b^b+a^a}}$$

The equality case is obvious .

The first reflex for me is to use rearrangement inequality but it gives just a little part of the inequality .The second reflex is to use power series of $x^x$ at $x=1$ . We get a polynomial and it's useful . Furthermore I have tried to denested the radical unsuccessfully.Finally my goal was to use with all of this the Buffalo's way but it's far .

Some remarks :

The inequality is very sharp because I think that we have $LHS-RHS\leq 10^{-2}$

Each coefficient under the root are $1$ behind the variable so maybe it's easier.

We have also :

Let $a,b,c>0$ such that $a+b+c=1$ and $a\ge b \geq c $ then we have :$$\frac{a}{a^a+b^b}+\frac{b}{b^b+c^c}+\frac{c}{c^c+a^a}\geq \frac{a}{a^a+c^c}+\frac{c}{c^c+b^b}+\frac{b}{b^b+a^a}$$

So the idea will be to put a power $2$ to each side and use rearrangement inequality to get other cases(maybe I have not checked that).

My last idea is in this link

If you have nice idea you are welcome .

Thanks a lot .

Answer 1

We will apply Ji Chen's Symmetric Function Theorem for $n=3$
(see https://artofproblemsolving.com/community/c6h194103p1065812):

Symmetric Function Theorem: Let $d\in (0,1)$. Let $x, y, z, u, v, w$ be non-negative real numbers satisfying $$x+y+z \ge u+v+w, \quad xy+yz+zx \ge uv+vw+wu, \quad xyz \ge uvw.$$ Then $x^d + y^d+z^d \ge u^d + v^d+w^d$.

Denote the six terms inside the root sign by $X, Y, Z, U, V, W$ respectively. Clearly, $XYZ = UVW$. We need to prove that $X+Y+Z\ge U+V+W$ and $XY+YZ+ZX \ge UV+VW+WU$.

1) To prove $X+Y+Z\ge U+V+W$, it suffices to prove that $$\frac{a-b}{a^a + b^b} + \frac{b-c}{b^b+c^c} \ge \frac{a-c}{c^c + a^a}$$ or $$\frac{a-b}{a^a + b^b} - \frac{a-b}{c^c + a^a} \ge \frac{b-c}{c^c + a^a} - \frac{b-c}{b^b+c^c}$$ or $$\frac{(a-b)(c^c - b^b)}{(a^a + b^b)(c^c + a^a)} \ge \frac{(b-c)(b^b-a^a)}{(c^c+a^a)(b^b+c^c)}$$ or $$(a-b)(c^{2c} - b^{2b}) \ge (b-c)(b^{2b} - a^{2a})$$ or $$\frac{b^{2b}-c^{2c}}{b-c} \le \frac{a^{2a}-b^{2b}}{a-b}.$$ This inequality is true since $a\ge b \ge c > 0$ and $x\mapsto x^{2x}$ is convex on $x > 0$.

2) To prove $XY+YZ+ZX \ge UV+VW+WU$, it suffices to prove that $$ab(a^a - b^b) + bc(b^b - c^c) + ca(c^c - a^a) \ge 0$$ or $$a^{a+1}(b-c) + c^{c+1}(a-b) \ge b^{b+1}(a-c)$$ or $$(a^{a+1}-b^{b+1})(b-c) \ge (b^{b+1} - c^{c+1})(a-b)$$ or $$\frac{a^{a+1}-b^{b+1}}{a-b} \ge \frac{b^{b+1} - c^{c+1}}{b-c}.\tag{1}$$ If $a\ge b\ge c > \mathrm{e}^{-3/2}$, the inequality is true since $x\mapsto x^{x+1}$ is convex on $x > \mathrm{e}^{-3/2} \approx 0.2231301602$.

If $c \le \mathrm{e}^{-3/2}$, (1) is written as $$\frac{b^b - c^c}{b-c} c \le \frac{a^a-b^b}{a-b} a.\tag{2}$$ Since $x\mapsto x^x$ is convex on $x > 0$, we have $$\frac{b^b - c^c}{b-c} \le \frac{a^a-b^b}{a-b}.$$ Thus, it suffices to prove that $a^a \ge b^b$. It is true. Omitted.