<s>[INST] Inequality dealing with area and sides of a triangle I have to prove that:
$$\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}\geq 8S^2$$
$a$, $b$, $c$ are the sides of a triangle, and $S$ its area.
I tried using the Sine Theorem, different  area formulas, and AM-GM but didn't get anywhere.
Thanks for your help!
 [/INST] From How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$, we can show that
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$
For your question, use Heron's formula:
$$S=\sqrt{s(s-a)(s-b)(s-c)}$$
$$s=\frac{a+b+c}{2}$$
So the right hand side of the inequality $8S^2$ can be turned into
$$8\left(\frac{a+b+c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)$$
which is equal to
$$\frac{1}{2}(a+b+c)(b+c-a)(a+c-b)(a+b-c)$$
Now, what's left is to prove
$$\frac {abc(a + b + c)}2\geq\frac{1}{2}(a+b+c)(b+c-a)(a+c-b)(a+b-c)$$
Or
$$abc\geq(b+c-a)(a+c-b)(a+b-c)$$,
which is true if we let $a\geq b\geq c$
So we are left with
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2\geq 8S^2$$
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