Proof of an inequality in arithmetic, geometric and harmonic means Prove that $$\left[\frac{x^2+y^2+z^2}{x+y+z}\right]^{x+y+z}\gt x^xy^yz^z\gt \left[\frac{x+y+z}{3}\right]^{x+y+z}$$
I could prove the inequality between the first two terms using the fact that $A.M\gt G.M$ in the following way. $$\left[\frac{x(\frac{x^2}{x})+y(\frac{y^2}{y})+z(\frac{z^2}{z})}{x+y+z}\right]\gt (x^xy^yz^z)^{\frac{1}{x+y+z}}$$
But I am unable to prove the inequality between the next two terms.

Just to spell out what Ivan says:
Consider the sequence $a_1,a_2,a_3 = x, y, z$ and weights $w_1,w_2,w_3 = x/s, y/s, z/s$, where $s  = x+y+z$. The inequality between harmonic and geometric (weighted) mean says that:
$$ \prod_{i} a_i^{w_i} \geq \frac{1}{\sum_i w_i \frac{1}{a_i}}.$$ 
Thus:
$$ x^{x/s}y^{y/s}z^{z/s} \geq \frac{1}{\frac{x}{s} \frac{1}{x} + \frac{y}{s} \frac{1}{y}+ \frac{z}{s} \frac{1}{z}} = \frac{1}{3 \cdot{1}{s} }= \frac{s}{3}.$$
Rising both sides to power $ s $ gives the claim:
$$ x^x y^y z^z \geq \left( \frac{x+y+z}{3} \right)^{x+y+z}.$$
