<s>[INST] Proof of an inequality involving different exponents 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?
 [/INST] I think the solution with derivative is very smooth:
$$f'(x)=\sum\limits_{cyc}\frac{a^x\ln a(b^x+c^x)-a^x(b^x\ln b+c^x\ln c)}{(b^x+c^x)^2}=$$
$$=\sum\limits_{cyc}\frac{a^x(b^x(\ln a-\ln b)-c^x(\ln c-\ln a))}{(b^x+c^x)^2}=$$
$$=\sum_{cyc}(\ln a-\ln b)\left(\frac{a^xb^x}{(b^x+c^x)^2}-\frac{a^xb^x}{(a^x+c^x)^2}\right)=$$
$$=\sum\limits_{cyc}\frac{a^xb^x(\ln a-\ln b)(a^x-b^x)(a^x+b^x+2c^x)}{(a^x+c^x)^2(b^x+c^x)^2}\geq0$$
</s>