geometric-inequalitiestriangles
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!
You still need to do this ~ C=sin-1(11/15) Because you are trying to find C alone. Correct me if I'm wrong.
From the Wikipedia article on Thales' theorem:
If you know the hypotenuse, you know the base of this diagram; if you also know the area, then you can compute the height (from $A=\frac12 bh$). Then you can draw a horizontal line at the required height; where it intersects the circle is the vertex with the right angle.
Best Answer
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 $$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2\geq 8S^2$$