if $x^y=y^x$ show that $x+y>2e$ 
Let $0<x<y$, such that
  $$x^y=y^x$$
  show that
  $$x+y>2e$$

Since $$y\ln{x}=x\ln{y}\Longrightarrow \dfrac{\ln{y}}{y}=\dfrac{\ln{x}}{x}$$
Let 
$$f(x)=\dfrac{\ln{x}}{x}\Longrightarrow f'(x)=\dfrac{1-\ln{x}}{x^2}$$
If $0<x<e$ then  $f'(x)>0$, if $x>e$,then $f'(x)<0$
$\Longleftrightarrow y-e+x-e>0$,maybe consider
$f(x-e)$ and $f(y-e)$,wihch bigger?
I just do it now 
Thanks in advance!
 A: 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 $g(r)$.
A: As  MathematicianByMistake answer,we want prove
$$g(r)=r^{\frac{1}{r-1}}+r^{\frac{r}{r-1}}=(x+1)^{\frac{1}{x}}+(x+1)^{\frac{x+1}{x}}=(x+1)^{\frac{1}{x}}(x+2),x>0$$
since
$$g'(x)=(x+1)^{\frac{1}{x}-1}\cdot\dfrac{x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}}{x^2}$$
we only prove
$$h(x)=x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}>0,x>0$$
since
$$h'(x)=3x(x+1)-(3x+2)\ln{(x+1)}>3x(x+1)-x(3x+2)=x>0$$
A: Alternative solution:
Fact 1: Let $f(u) = \frac{\ln u}{u}$. Then, $f(u)$ is strictly increasing on $(0, \mathrm{e})$, and strictly decreasing on $(\mathrm{e}, \infty)$.
(The proof is easy and thus omitted.)
By Fact 1, it is easy to prove that $1 < x < \mathrm{e} < y$.
We need to prove that $y > 2\mathrm{e} - x$.
Since $2\mathrm{e} - x \in (\mathrm{e}, \infty)$, by Fact 1, it suffices to prove that
$f(y) < f(2\mathrm{e} - x)$.
Since $f(x) = f(y)$, it suffices to prove that
$f(x) < f(2\mathrm{e} - x)$ that is
$$\frac{\ln x}{x} < \frac{\ln (2\mathrm{e} - x)}{2\mathrm{e} - x}.$$
It suffices to prove that, for all $x$ in $(1, \mathrm{e})$,
$$x\ln (2\mathrm{e} - x) - (2\mathrm{e} - x)\ln x > 0. $$
Denote LHS by $g(x)$. We have
$$g'(x) = \ln (2\mathrm{e} - x) - \frac{x}{2\mathrm{e} - x} + \ln x - \frac{2\mathrm{e} - x}{x}$$
and
$$g''(x) = \frac{2(\mathrm{e} - x)(4\mathrm{e}^2 + 2\mathrm{e} x - x^2)}{(2\mathrm{e} - x)^2x^2}.$$
Since $g''(x) > 0$ for all $x$ in $(1, \mathrm{e})$
and $g'(\mathrm{e}) = 0$, we have $g'(x) < 0$ for all $x$ in $(1, \mathrm{e})$.
Since $g(\mathrm{e}) = 0$, we have $g(x) > 0$ for all $x$ in $(1, \mathrm{e})$.
We are done.
See also:
Prove $(x-1)(y-1)>(e-1)^2$ where $x^y=y^x$, $y>x>0$.
