About the absolute inequality of variables $x,y,z$ How to prove that $$e^{x}(2x-y-z)+e^{y}(2y-x-z)+e^{z}(2z-x-y)\geq 0$$ for all $x,y,z\in\mathbb{R}$.
 A: You can use Rearrangement inequality. Without losing the generality $x \leq y \leq z$, so $e^x \leq e^y \leq e^z$, so by Rearrangement inequality:
$$xe^x+ye^y+ze^z \geq ye^x+ze^y+xe^z$$
and
$$xe^x+ye^y+ze^z \geq ze^x+xe^y+ye^z$$
If you add this inequalities side by side you get your inqeuality.
A: After rearrangement, your inequality simplifies to
$$
3(xe^x+ye^y+ze^z)\geq(x+y+z)(e^x+e^y+e^z)
$$
or, even more suggestively,
$$
\frac{1}{3}(xe^x+ye^y+ze^z)\geq\frac{1}{3}(x+y+z)\frac{1}{3}(e^x+e^y+e^z).\tag{*}
$$
Because $\log$ is an increasing transformation, $e^x\geq e^y$ iff $x\geq y$, (*) follows immediately from the Chebyshev's sum inequality.
A: Note that for $u\in \mathbb{R}$ we have $u(\exp(u)-1)\geq 0$. Now replace $u$ by $\displaystyle \frac{2x-y-z}{3}$, $\displaystyle \frac{2y-x-z}{3}$, $\displaystyle \frac{2z-y-x}{3}$, add the $3$ inequalities. The constant term is zero;  and now multiply by $\displaystyle 3\exp((x+y+z)/3)$.   
A: Write
$$a=2x-y-z\ ,\quad b=2y-x-z\ ,\quad c=2z-x-y$$
and note that
$$a+b+c=0\ ,$$
so that $a,b,c$ cannot all be negative.
If none of them is negative then they must all be zero and the result is obvious.
If one of them is negative, by symmetry we may assume it is $c$.  Then we have
$$0\le\frac{2a+b}{3}=x-z\ ,\quad 0\le\frac{a+2b}{3}=y-z$$
and therefore
$$e^xa+e^yb+e^zc=(e^x-e^z)a+(e^y-e^z)b\ge0\ .$$
If two of $a,b,c$ are negative, by symmetry assume they are $b$ and $c$; by a similar argument we get
$$x>y\ ,\quad x>z$$
and so
$$e^xa+e^yb+e^zc=(e^x-e^y)(-b)+(e^x-e^z)(-c)\ge0\ .$$
