TITLE: How to prove this inequality $\frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}\ge 3$
QUESTION [16 upvotes]: 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!

REPLY [2 votes]: 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 $y$.