Prove that $\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \ge \frac{(a+b+c)^3}{3(x+y+z)}$ a,b,c,x,y,z are positive real numbers. I stumbled upon it on some olympiad papers. Tried to AM>GM but didn't get any idea to move forward.

Holder's Inequality gives:
$$\left( \frac{a^3}x + \frac{b^3}y + \frac{c^3}z \right)(x + y + z)(1+1+1) \ge (a+b+c)^3 $$
