if $x^y=y^x$ show that $x+y>2e$

We can see that $$f(x)=\frac{\ln x}{x}$$ is strictly increasing in $(0,e)$ and striclty decreasing in $(e,+\infty)$.

Thus the equality $f(x)=f(y)$ cannot hold for the case of $x,y \in (0,e)$ or $x,y \in (0,+\infty)$.

Since $x\lt y$, it follows that $x\in (0,e)$ and $y \in (e,+\infty)$.

So let $y=rx$, for $r\in (1,+\infty)$ and we obtain $$(xr)^x = (x)^{rx}\implies rx \ln x = x (\ln r + \ln x)\implies(r−1)\ln x = \ln r$$

So, $$x = r^{1/(r−1)}, y= r^{r/(r−1)}$$ and we need to show that for $r\in (1,+\infty)$, the function $$g(r)=r^{1/(r−1)}+r^{r/(r−1)}\gt2e$$

We notice that $$\lim_{r\rightarrow1}r^{1/(r−1)}=\lim_{r\rightarrow1}r^{r/(r−1)}=e$$

Indeed, if we set $u=\frac1r$ we get $\lim(1+\frac1u)^u=e$, and similarly for the second limit.

Thus we only need to show that $g(r)$ is strictly increasing in $(1,+\infty)$.

EDIT
(As clark rightly pointed, the first approach to prove that $g(r)$ is strictly increasing was wrong on my part-see the comments below. I hope this new take is correct.)

We wish to show that $g(r)=r^{\frac{1}{r-1}}(r+1)$ is strictly increasing. We will show that $h(r)=\ln g(r)$ is strictly increasing and since $\ln x$ is s.i, we derive from this the fact that $g(r)$ is s.i. (That is, if $h(x)=\ln g(x)$ is strictly increasing $\Rightarrow g(x)$ is strictly increasing. To show this one can simply use Reductio ad absurdum and assume $h(x)$ is s.i while $g(x)$ is not s.i. )

So, we have $$\ln g(r)=\ln r^{\frac{1}{r-1}}(r+1)=\ln r^{\frac{1}{r-1}}+\ln(r+1)=\frac1{r-1}\ln r+\ln (r+1)\\h'(r)=\ln r+\frac{1}{r(r-1)}+\frac1{r+1}\gt0, \forall r \in (1,+\infty)$$ It follows that $h(r)$ is strictly increasing and so is $g(r)$.