I was asked the following interesting question but I can not come up with a proof.
Let $a, b, c$ be positive real numbers. Prove that $$ \frac{a^3}{b^3 + c^3} + \frac{b^3}{a^3 + c^3} + \frac{c^3}{a^3 + b^3} \geq \frac{a^2}{b^2 + c^2} + \frac{b^2}{a^2 + c^2} + \frac{c^2}{a^2 + b^2} \tag{$\clubsuit$} $$
One way to prove $(\clubsuit)$ may be to consider the following function $$ f(x) = \frac{a^x}{b^x + c^x} + \frac{b^x}{a^x + c^x} + \frac{c^x}{a^x + b^x} $$ and prove that $f(x)$ is a non-decreasing function on $[2, 3]$. When we take its derivative, the formula looks quite tedious and we just can not continue on. So is there any better solution?
I think the solution with derivative is very smooth: $$f'(x)=\sum\limits_{cyc}\frac{a^x\ln a(b^x+c^x)-a^x(b^x\ln b+c^x\ln c)}{(b^x+c^x)^2}=$$ $$=\sum\limits_{cyc}\frac{a^x(b^x(\ln a-\ln b)-c^x(\ln c-\ln a))}{(b^x+c^x)^2}=$$ $$=\sum_{cyc}(\ln a-\ln b)\left(\frac{a^xb^x}{(b^x+c^x)^2}-\frac{a^xb^x}{(a^x+c^x)^2}\right)=$$ $$=\sum\limits_{cyc}\frac{a^xb^x(\ln a-\ln b)(a^x-b^x)(a^x+b^x+2c^x)}{(a^x+c^x)^2(b^x+c^x)^2}\geq0$$