For positive real numbers $x,y,z$, Prove that 
For positive real numbers $x,y,z$, Prove that,
$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z \geq
 \bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$

I have no idea how to begin this? I am getting a hint  for A.M. - G.M. equality but numbers don't seem to fit.
Thanks for you time.
 A: Since $\ln$ is a concave function we obtain:
$$\sum_{cyc}\frac{x}{x+y+z}\ln{x}\leq\ln\left(\sum_{cyc}\frac{x}{x+y+z}\cdot x\right)=\ln\frac{x^2+y^2+z^2}{x+y+z},$$
which gives a left inequality.
The right inequality it's just Jensen for the convex function $f(x)=x\ln{x}:$
$$\frac{\sum\limits_{cyc}x\ln{x}}{3}\geq\frac{x+y+z}{3}\ln\frac{x+y+z}{3}.$$
A: HINT:

Try using $x$ times $x$, $y$ times $y$ , $z$ times $z$.

EDIT:

Seemingly people downvote hints so updating the full solution.


Using A.M. - G.M. inequality,
$$\frac{x+x+..(x \;\Bbb{times})+y+y+..(y \;\Bbb{times})+z+z+..(z \;\Bbb{times})}{x+y+z}\geq (x^xy^yz^z)^{\frac{1}{x+y+Z}}$$
$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z $$
Using G.M. - H.M. inequality,
$$(x^xy^yz^z)^{\frac{1}{x+y+Z}} \geq \frac{x+y+z}{\frac{1}{x}+\frac{1}{x}+..(x \; \Bbb{times})+\frac{1}{y}+\frac{1}{y}+..(y \; \Bbb{times})+\frac{1}{z}+\frac{1}{z}+..(z \; \Bbb{times})}$$
$$ x^xy^yz^z \geq
 \bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$
$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z \geq
 \bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$
