Prove that $\frac{a^6}{(2b+5c)^2}+\frac{b^6}{(2c+5a)^2}+\frac{c^6}{(2a+5b)^2}\geq \frac{3}{49}$. I am struggling to find the solution for this problem:

Problem. If $a,b,c>0$ and $a^4+b^4+c^4\geq3$, prove that
$$\frac{a^6}{(2b+5c)^2}+\frac{b^6}{(2c+5a)^2}+\frac{c^6}{(2a+5b)^2}\geq \frac{3}{49}.$$

I see the minimum can be found for $a=b=c=1$. However the usual methods attacking this inequality don't get me very far.
Any help is more than welcome
 A: As suggested by River Li, I think I found a proof. Happy to have any comment:

First take $\lambda = \sqrt[4]{\frac{3}{a^4+b^4+c^4}}<=1 $

Using $a'=\lambda a$,$b'=\lambda b$ and $c'= \lambda c$, we have $a',b',c'>0$ , $a'^{4}+b'^{4}+c'^{4}= 3$ and the inequality to prove become:
$$ \frac{a'^6}{(2b'+5c')^2} +\frac{b'^6}{(2c'+5a')^2}+\frac{c'^6}{(2a'+5b')^2} \ge \lambda^4\frac{3}{49} $$
As $\lambda^4 \le 1$, we only need to show:
$$ \frac{a'^6}{(2b'+5c')^2} +\frac{b'^6}{(2c'+5a')^2}+\frac{c'^6}{(2a'+5b')^2} \ge \frac{3}{49} $$
Then WLOG, we can assume that $a^4+b^4+c^4 = 3$

Now using mean inequalities, we have:
$$ \frac{a^2+b^2+c^2}{3}\le \sqrt{\frac{a^4+b^4+c^4}{3}}\le 1 $$
Then
$$\frac{a+b+c}{3}\le \sqrt{\frac{a^2+b^2+c^2}{3}} \le 1$$
And
$$ abc \le (\frac{a+b+c}{3})^3 \le 1 $$
Furthermore, we have using C-S :
$$ \sum_{cycl}^{} a^2b^2 \le \sum_{cycl}^{} a^4 \le 3$$
Using again C-S, we have:
$$(\sum_{cycl}^{} a^4)^2 \le \sum_{cycl}^{}\frac{a^6}{(2b+5c)^2}\sum_{cycl}^{}a^2(2b+5c)^2$$
or
$$ \frac{9}{\sum_{cycl}^{}a^2(2b+5c)^2} \le \sum_{cycl}^{}\frac{a^6}{(2b+5c)^2} $$
But
$\sum_{cycl}^{}a^2(2b+5c)^2= 20abc(a+b+c) + 29\sum_{cycl}^{} a^2b^2 \le 20 \times 3 + 29 \times 3 = 49 \times 3 $

Thus
$$ \frac{3}{49} \le \sum_{cycl}^{}\frac{a^6}{(2b+5c)^2} $$
