Question: Let $x>1$, $y>1$ and $x\neq y$ such that $x^y=y^x$. Prove that:
$$x^y>e^e$$

We can assume $y=kx$, where $k>1$. 

Hence, $x=k^{\frac{1}{k-1}}$ and $y=k^{\frac{k}{k-1}}$ and we need to prove that $f(k)>e^e$, where $f(k)=k^{\frac{k^{\frac{k}{k-1}}}{k-1}}$.

We can show that $f'(k)>0$ for $k>1$ and $\lim\limits\_{k\rightarrow1^+}f(k)=e^e$ and we are done.

Is there a nice proof for this inequality? Thank you!

Answer: 
We have to minimize the function $$f(x)=x^{y}\tag{1}$$ where $x$ and $y$ are connected by equation $$x^{y}=y^{x}\tag{2}$$ As usual we need to find derivative $f'(x)$ and this is bit tricky here. First we take logarithm of the equation $(2)$ to get $$y\log x = x\log y$$ and then differentiating with respect to $x$ we get $$\frac{y}{x} + y'\log x = \log y + \frac{xy'}{y}$$ or $$y' = \frac{y}{x}\cdot\frac{y - x\log y}{x - y\log x}\tag{3}$$ From $(1)$ we get $$\log f(x) = y\log x$$ and differentiation gives us $$\frac{f'(x)}{f(x)} = \frac{y}{x} + y'\log x$$ and using equation $(3)$ we get $$\frac{f'(x)}{f(x)} = \frac{y}{x}\left(1 + \frac{y\log x - x\log x\log y}{x - y\log x}\right)$$ and thus $f'(x) = 0$ is possible only when $$\log x\log y = 1\tag{4}$$ so that $y = \exp(1/\log x)$ and $$f(x) = x^{y} = y^{x} = \exp\left(\frac{x}{\log x}\right)$$ It can be easily checked that the function $g(x) = x/\log x$ attains it's minimum value at $x = e$ and therefore $f(x) = e^{g(x)}$ also attains it's minimum value at $x = e$ and therefore the minimum value of $f(x) = x^{y}$ is $e^{e}$.
I count this as a partial answer: If you keep to the branch $y \le x$ (which given the symmetry of the problem I think is OK), then using the *known* properties of the Lambert function you have $y = x$ for $0 \le x \le e$. Now consider $x^y$. For $0 \le x \le e$ you have $x^y = x^x$. This solution can be considered "known" in the sense that elementary means can used to deduce its properties, such as general shape. If you know consider the equation $y = -x \frac{W(-\frac{\ln x}{x})}{\ln x}$ and look at $x^y$ then Maple gives the following far from obvious structure (see attached figure).

To complete the answer you would have to show that $\frac{d x^y}{d x} \ge 0$ for $x \ge e$. This looks a fearsome task but it may be possible using only known properties of Lambert's function (for example) $\frac{d W(z)}{d z} = \frac{W(z)}{z (1 + W(z))}$ (from Wiki).

From the figure you seem to have (for the chosen branch) that $x^y > e^e$ for $x > e$.
[![enter image description here](https://i.stack.imgur.com/VMGe8.png)](https://i.stack.imgur.com/VMGe8.png)