Inequality : $\sum_{cyc}\Big(\frac{x^4}{8x^3+5y^3}\Big)^3\geq \frac{x^3+y^3+z^3}{13^3}$

Question

It's related to this Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$:

$x,y,z >0$, prove $$\Big(\frac{x^4}{8x^3+5y^3}\Big)^3+\Big(\frac{y^4}{8y^3+5z^3}\Big)^3+\Big(\frac{z^4}{8z^3+5x^3}\Big)^3 \geqslant \frac{x^3+y^3+z^3}{13^3}$$

The inequality is equivalent to : $$x^3\Big(\frac{1}{8+5(\frac{y}{x})^3}\Big)^3+y^3\Big(\frac{1}{8+5(\frac{z}{y})^3}\Big)^3+z^3\Big(\frac{1}{8+5(\frac{x}{z})^3}\Big)^3 \geqslant \frac{x^3+y^3+z^3}{13^3}$$

Or :

$$\frac{x^3\Big(\frac{1}{8+5(\frac{y}{x})^3}\Big)^3+y^3\Big(\frac{1}{8+5(\frac{z}{y})^3}\Big)^3+z^3\Big(\frac{1}{8+5(\frac{x}{z})^3}\Big)^3}{x^3+y^3+z^3} \geqslant \frac{1}{13^3}$$

Now we apply the Jensen's inequality to the function $f(x)=\frac{1}{(8+5x)^3}$ wich is clearly convex we get :

$$\frac{x^3 f((\frac{y}{x})^3)+y^3 f((\frac{z}{y})^3)+z^3 f((\frac{x}{z})^3)}{x^3+y^3+z^3} \geqslant f\Big(\frac{x^3 (\frac{y}{x})^3+y^3 (\frac{z}{y})^3+z^3 (\frac{x}{z})^3}{x^3+y^3+z^3}\Big)=\frac{1}{13^3}$$

And we are done .

My question is have you an nice alternative proof ?

Thanks a lot for your time .

Answer 1

By Holder $$\sum_{cyc}\left(\frac{x^4}{8x^3+5y^3}\right)^3\left(\sum_{cyc}(8x^3+5y^3)\right)^3\geq(x^3+y^3+z^3)^4.$$ Thus, it's enough to prove that $$(x^3+y^3+z^3)^4\geq\frac{x^3+y^3+z^3}{13^3}\cdot\left(\sum_{cyc}(8x^3+5y^3)\right)^3,$$ which is an equality.