<s>[INST] 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!
 [/INST] 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. 
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