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 $e^{e}$.
You certainly have $\_{k \to 1^+}k^{\frac{k}{k-1}} = e$. This follows by considering $z = k^{\frac{k}{k-1}}$ so that $\ln z = \frac{k}{k-1} \ln k$, then lim$\_{k \to 1^+} \ln z = \frac{1 + \ln k}{1} = 1$ (L'Hopital) *i.e.* $z = e$ (alternatively, look at the definition of $e = $ lim$\_{n \to \infty} (1 + \frac{1}{n})^n$). Then look at $f(k) = z^z$ and write $\ln(f) = z \ln z$ to show lim$\_{k \to 1^+} \ln(f) \to e \ln e$, giving lim$\_{k \to 1^+} f(k) = e^e$.