How to prove this inequality $\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}\ge 3$ 
let $x,y,z$ be positive numbers, and such $x+y+z=1$

show that
$$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}\ge 3$$
My try:

let
  $$a=\ln{\dfrac{x^y}{y^x}},b=\ln{\dfrac{y^z}{z^y}},c=\ln{\dfrac{z^x}{x^z}}$$
  so
  $$a=y\ln{x}-x\ln{y},b=z\ln{y}-y\ln{z},c=x\ln{z}-z\ln{x}$$
  and we note 
  $$az+bx+yc=(y\ln{x}-x\ln{y})z+(z\ln{y}-y\ln{z})x+(x\ln{z}-z\ln{x})y=0$$
  so
  $$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}=e^a+e^b+e^c$$
  so
  $$\Longleftrightarrow e^a+e^b+e^c\ge 3$$
  But then I can't prove it.

If this problem is to prove
$$ze^a+xe^b+ye^c\ge 3,$$
I can prove it,because 
$$ze^a+xe^b+ye^c\ge=\dfrac{z}{x+y+z}e^a+\dfrac{x}{x+y+z}e^b+\dfrac{y}{x+y+z}e^c$$
so
use Jensen's inequality,we have
$$ze^a+xe^b+ye^c\ge e^{\dfrac{az+bx+yc}{x+y+z}}=3$$
This problem comes from How prove this $\dfrac{x^y}{y^x}\ge (1+\ln{3})x-(1+\ln{3})y+1$?
Thank you very much!

Another solution.
Without loss of generality, we can assume that $0<x\le y\le z$ and so $x=az$, $y=bz$, $0<a\le b\le1$. By the AM-GM inequality, we find that
\begin{align}
\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}&\ge 3\sqrt[3]{x^{y-z}y^{z-x}z^{x-y}}=3\sqrt[3]{(az)^{(b-1)z}(bz)^{(1-a)z}z^{(a-b)z}}=3\sqrt[3]{a^{(b-1)z}b^{(1-a)z}}=\\
&=3(a^{b-1}b^{1-a})^{z/3}
\end{align}
so we only need to prove that $a^{b-1}b^{1-a}\ge1$, which Hansen did in the answer below. We also see that the condition $x+y+z=1$ is not needed.
