Olympiad inequalities. I am trying to solve this problem but unable to. Can someone please give some hint or help. 
I have to use holder inequality.  
For $a,b,c$ positive real numbers prove. 
$ \frac{a^6}{b^2+c^2} + \frac{b^6}{c^2+a^2} + \frac{c^6}{a^2+b^2} \ge \frac{abc(a+b+c)}{2}$

By Cauchy-Schwarz inequality (Titu's lemma or Sedrakyan's inequality specifically):
$$F=\frac{a^6}{b^2+c^2}+\frac{b^6}{c^2+a^2}+\frac{c^6}{a^2+b^2} \ge 
\frac{(a^3+b^3+c^3)^2}{2(a^2+b^2+c^2)}$$
Next use, $$(a^3+b^3+c^3) \ge \frac {(a^2+b^2+c^2)(a+b+c)}{3}~~\text{(Tchebechev)}$$
So, 
$$F \ge \frac{1}{18} \ \underbrace{(a^2+b^2+c^2)}_\text{$AM \ge GM$} \ \underbrace{(a+b+c)}_\text{$AM \ge GM $} (a+b+c)$$
$$ F \ge \frac{1}{18} \left(3(abc)^{\frac{2}{3}}\cdot 3(abc)^{\frac{1}{3}}\cdot(a+b+c)\right) = \frac{abc(a+b+c)}{2}. $$
