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?
 A: If you know that
$$x^y=y^x \implies x=-\frac{y}{\log (y)}W\left(-\frac{\log (y)}{y}\right)$$ where $W(t)$ is Lambert function, then
$$(x-1)(y-1)=(1-y)\Bigg[1+ \frac{y}{\log (y)}W\left(-\frac{\log (y)}{y}\right)\Bigg]=f(y)$$
$f'(y)$ cancels at $y=e$ and, at this point, $x=e$. The second derivative $$f''(e)=\frac{2 (2 e-5)}{3 e}\sim 0.107069~~ >~~ 0$$ confirms that this point is a minimum and $f(e)=(e-1)^2$.
Edit
As @Martin R commented, I did not prove that this is the global minimum.
In fact, this can be done using the fact that $\forall t \geq e$
$$W(t) \geq\log (t)-\log (\log (t))+\frac 12\frac{\log (\log (t))}{ \log (t)}=f(t)$$
$$W(t) \leq \log (t)-\log (\log (t))+\frac e{e-1}\frac{\log (\log (t))}{ \log (t)}=g(t)$$ or even better tha approximation of $W(t)$ I gave here.
A: Continuing your idea, let $a = 1+ \dfrac 1t, t > 0.$ Then, you can write your inequality as $xy-x-y > e^2-2e$, which is in terms of the variable $t:$
$$f(t) = t\ln\left(1+\frac 1t\right) + \ln\left(\left(1+\frac 1t\right)^{t+1}-2-\frac 1t\right) > 1+\ln(e-2).$$
Wolfram says that this is indeed a decreasing function in $t$, so the infimum will be:
$$\lim\limits_{t\to\infty} f(t) = 1+ \ln(e-2).$$
Then, the only thing to show now is the fact that $f(t)$ is decreasing, but I suspect this will take an ugly computation.
EDIT: Actually, the derivative did not turn out to be too bad:
\begin{align}f'(t) &= \\
&\dfrac{\dfrac{1}{t^2} + \left(1+\dfrac 1t\right)^{t+1}{\left(2\ln\left(1+\frac 1t\right)-\dfrac 1t-\dfrac{1}{t+1}\right)} + \left(2+\dfrac{1}{t}\right)
\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\left(1+\dfrac 1t\right)^{t+1} - \dfrac 1t-2} &
\\<0\end{align}
Finally, the above is equivalent to:
$$\left(1+\dfrac 1t\right)^{t+1} > \dfrac{\dfrac{1}{t^2} + \left(2+\dfrac 1t\right)\left(\dfrac{1}{1+t} - \ln\left(1+\dfrac{1}{t}\right)\right)}{\dfrac{1}{t}+\dfrac{1}{1+t} - 2\ln\left(1+\dfrac{1}{t}\right)},$$
which I think should be doable by brute-force expanding with a Taylor's series etc.
A: (A comment rather than an answer
but much easier to enter)
This is an improvement
of my answer
Proving or disproving: $a^b<b^a$?
where I showed that
if $x^{1/x} = y^{1/y}$
and
$1 < x < y$
then
$y > e^2/x$.
This is an improvement since
$(e-1)^2/(x-1)+1- e^2/x
=(e - x)^2/((x - 1) x)
\gt 0$.
A: 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 $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$
