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.

Macavity's way is certainly the most elegant. Here is a way to do it just using Cauchy-Schwarz and AM-GM. Use Cauchy-Schwarz inequality to get:
$$(x+y+z)\left(\frac{a^3}{x}+\frac{b^3}{y}+\frac{c^3}{z}\right)\geq (a^{3/2}+b^{3/2}+c^{3/2})^2$$
So to prove your inequality, it suffices to show that
$$ (a^{3/2}+b^{3/2}+c^{3/2})^2 \geq \frac{(a+b+c)^3}{3}$$
which can be proved as follows. It is equivalent to (after expanding):
$$3(a^3+b^3+c^3)+6(a^{3/2}b^{3/2}+b^{3/2}c^{3/2}+c^{3/2}a^{3/2})\geq a^3+b^3+c^3 +3ab(a+b)+3bc(b+c)+3ca(c+a)+6abc$$
In other words, we want to prove:
$$ 2(a^3+b^3+c^3)+6(a^{3/2}b^{3/2}+b^{3/2}c^{3/2}+c^{3/2}a^{3/2})\geq 3ab(a+b)+3bc(b+c)+3ca(c+a)+6abc$$
Now apply AM-GM:
$$a^3+a^{3/2}b^{3/2}+a^{3/2}b^{3/2}\geq 3a^{2}b$$
$$b^3+a^{3/2}b^{3/2}+a^{3/2}b^{3/2}\geq 3ab^2$$
$$b^3+b^{3/2}c^{3/2}+b^{3/2}c^{3/2}\geq 3b^{2}c$$
$$c^3+b^{3/2}c^{3/2}+b^{3/2}c^{3/2}\geq 3bc^2$$
$$c^3+c^{3/2}a^{3/2}+c^{3/2}a^{3/2}\geq 3c^{2}a$$
$$a^3+c^{3/2}a^{3/2}+c^{3/2}a^{3/2}\geq 3ca^2$$
$$2(a^{3/2}b^{3/2}+b^{3/2}c^{3/2}+c^{3/2}a^{3/2})\geq 6abc$$
Now add all of these seven inequalities to get the desired inequality.
