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!
 A: This is to prove $a^{b-1}b^{1-a}\ge 1$ as needed in Ron Ford's answer above. Let $f(b)=(1-b)(-\log a)-(1-a)(-\log b)$, $b\in [a,1]$. $f(a)=f(1)=0$. $f$ is concave, as $f''(b)=-\frac{1-a}{b^2}<0$. So $f(b)>0,\,\forall b\in(a,1)$ and the result follows. 
A: The inequality as posed in the OP appears to be correct.  The accepted proof (Ford/Hansen) is not.  
Note that the proof doesn't require the assumption that $x+y+z=1$.  Take $x=6$, $y=2$, and $z=1$.  Then the left side is ${36\over 64} + 2 + {1\over 6} =  2 + {35\over 48} < 3$, violating the inequality.  Clearly the proof cannot be correct as stated.
The error in the Ford/Hansen proof is the assumption we can take $x \leq y \leq z$.  This is not generally possible because the LHS of the inequality is not invariant to permuting any pair of the variables.  Note that if we swap $x$ with $y$, the first term becomes its own reciprocal and the second and third terms become each other's reciprocals:
Define 
$$A = x^{y}/y^{x}, B = y^{z}/z^{y}, C = z^{x}/x^{z}.$$
Exchanging $x$ with $y$ makes $A \rightarrow 1/A$, $B \rightarrow 1/C$, and $C \rightarrow 1/B$.  
Furthermore, no proof using the AM-GM inequality can work.  Applying AM-GM we get:
$$ A + B + C \geq 3(ABC)^{1/3}.$$
If $ABC < 1$, we can't finish the proof.  If $ABC > 1$, then consider the case where we swap $x$ and $y$:
$$ A^{-1} + B^{-1} + C^{-1} \geq 3(ABC)^{-1/3}$$
Unless $ABC=1$, the bound is insufficient in one of the two cases above to prove the desired result. If the bound works for a given $x$, $y$, and $z$ then it won't suffice for the case where we swap $x$ and $y$.
A: 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.
