Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$. 
Let $x,y$ be different positive real numbers satisfying $x^y=y^x$.
Prove $(x-1)(y-1)>(e-1)^2$.

We may suppose $a=\frac{y}{x}>1$. Then we obtain $x=a^{\frac{1}{a-1}},y=a^{\frac{a}{a-1}}.$ But how to go on?

Sketch of a proof:
Fact 1: Let $f(u) = \frac{\ln u}{u}$. Then, $f(u)$ is strictly increasing on $(0, \mathrm{e})$,
and strictly decreasing on $(\mathrm{e}, \infty)$.
WLOG, assume $y < x$. From $x^y = y^x$, we have $\frac{\ln x}{x} = \frac{\ln y}{y}$.
By Fact 1, it is easy to prove that $1 < y < \mathrm{e} < x$.
We need to prove that $y > 1 + \frac{(\mathrm{e} - 1)^2}{x-1} \triangleq y_0$. Clearly, $y_0\in (1, \mathrm{e})$.
By Fact 1, it suffices to prove that $f(y) > f(y_0)$
or $\frac{\ln x}{x} > f(y_0)$. It suffices to prove that
$$\frac{\ln x}{x} > \frac{\ln\left( 1 + \frac{(\mathrm{e} - 1)^2}{x-1}\right)}{1 + \frac{(\mathrm{e} - 1)^2}{x-1}}, \ \forall x > \mathrm{e}.$$
It is true. But my proof is not nice. (Hint: Use Fact 2, then use derivative.)
Fact 2: $\frac{\ln v}{v} \ge \frac{(2\mathrm{e} + 10)v - 16\mathrm{e} + 4\mathrm{e}^2}
{(3\mathrm{e} - 3)v^2 + (16\mathrm{e} - 4\mathrm{e}^2)v - 19\mathrm{e}^2 + 7\mathrm{e}^3}$ for all $v \ge \mathrm{e}$.
(Hint: Take derivative.)
Remark 1: I propose the following problem:
Let $0 < y < x$ such that $\frac{\ln x}{x} = \frac{\ln y}{y} = a$ ($a>0$). Prove that
$$(x-1)(y-1) > \left(\mathrm{e} - 1 + \frac{5}{8}\left(\frac{1}{\mathrm{e}} - a\right)\right)^2.$$
By the way, there are many similar problems in MSE or AoPS. These problems has the following description:
Let $f(z)$ be a unimodal function. Let $f(x) = f(y) = a$. Then $g(x, y) \ge h(a)$.
See my closed post:
Inequalities involving roots of some functions (e.g., $\frac{\ln x}{x}$, $x\ln x$)
For example,
Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$,
let $f(x) = (x-1)\ln x$, and given $0 < a < b$. If $f(a) = f(b)$, prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$
