Inequality triangle Radon substitutions

Question

I have this inequality: $$\sum \frac {a^3}{p-a}\geq 8(2R-r)^2$$ I have tried using Radon substitutions and I get this: $$\sum \frac{(y+x)^3}{x}\geq 8(2R-r)^2$$ I know from Holder that : $$\sum \frac{(y+x)^3}{x}\geq \frac{(2(x+y+z))^3}{3(x+y+z)}=\frac{8(x+y+z)^2}{3}$$ Now I want to prove that: $$\frac{8(x+y+z)^2}{3}\geq 8(2R-r)^2$$ And after that I get: $$p^2\geq 48(2R-r)^2$$ Now here I'm not sure that I can prove this. If someone has a hint or some help I would apreciate.

Answer 1

In the standard notation we need to prove that $$\sum_{cyc}\frac{a^3}{b+c-a}\geq4\left(\frac{abc}{2S}-\frac{2S}{a+b+c}\right)^2.$$ Now, let $a=y+z$, $b=x+z$ and $c=x+y$.

Thus, $x$, $y$ and $z$ are positives and we need to prove that $$\sum_{cyc}\frac{(y+z)^3}{2x}\geq\frac{(abc(a+b+c)-4S^2)^2}{S^2(a+b+c)^2}$$ or $$\sum_{cyc}\frac{(y+z)^3}{2x}\geq\frac{(2(x+y)(x+z)(y+z)(x+y+z)-4xyz(x+y+z))^2}{4xyz(x+y+z)^3}$$ or $$\sum_{cyc}\frac{(y+z)^3}{2x}\geq\frac{((x+y)(x+z)(y+z)-2xyz)^2}{xyz(x+y+z)}$$ or $$(x+y+z)\sum_{cyc}xy(x+y)^3\geq2\left(\sum_{cyc}(x^2y+x^2z)\right)^2.$$ Now, by C-S $$(x+y+z)\sum_{cyc}xy(x+y)^3=\frac{1}{2}\sum_{cyc}(x+y)\sum_{cyc}xy(x+y)^3\geq\frac{1}{2}\left(\sum_{cyc}\sqrt{xy}(x+y)^2\right)^2.$$ Thus, it's enough to prove that $$\sum_{cyc}\sqrt{xy}(x+y)^2\geq2\sum_{cyc}(x^2y+xy^2)$$ or $$\sum_{cyc}\sqrt{xy}(x+y)^2\geq2\sum_{cyc}xy(x+y)$$ or $$\sum_{cyc}\sqrt{xy}(x+y)(\sqrt{x}-\sqrt{y})^2\geq0.$$