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 think this can be solved by recasting it in the language of Lagrange multipliers. We wish to maximize the function $f(x,y,z) = z$ subject to the constraints $g\_1(x,y,z) = x^y - z= 0$ and $g\_2(x,y,z) = y^x - z = 0$. By the properties of Lagrange multipliers, $f$ will have a local extremum on the surface specified by these constraints if there exists a point where
$$
\nabla f + \lambda\_1 \nabla g\_1 + \lambda\_2 \nabla g\_2 = 0.
$$
Taking these gradients, we find that we must have
\begin{align\*}
\lambda\_1 [y x^{y-1}] + \lambda\_2 [(\ln y) y^x] &= 0 \\
\lambda\_1 [(\ln x) x^y] + \lambda\_2 [x y^{x-1}] &= 0 \\
1 -\lambda\_1 - \lambda\_2 &= 0
\end{align\*} 
We can then eliminate $\lambda\_2$ from these equations:
\begin{align\*}
\lambda\_1 [y x^{y-1}] + (1 - \lambda\_1) [(\ln y) y^x] &= 0 \\
\lambda\_1 [(\ln x) x^y] + (1 - \lambda\_1) [x y^{x-1}] &= 0
\end{align\*} 
Then we can solve each equation for $\lambda\_1$ to eliminate it from our system:
$$
\lambda\_1 = \frac{(\ln y) y^x}{(\ln y) y^x - y x^{y-1}} = \frac{x y^{x-1}}{x y^{x-1} - (\ln x) x^y } 
$$
Inverting both sides of this equation and simplifying yields
$$
\frac{y x^{y-1}}{(\ln y) y^x} = \frac{(\ln x) x^y }{x y^{x-1}},
$$
which further simplifies to
$$
(\ln x) (\ln y) = 1.
$$
Thus, at the extremum (if it exists), we must have $y = e^{1/\ln x}$ and $z = y^x = e^{x/\ln x}.$ But on the interval $(1, \infty)$, the function $x / \ln x$ has a global minimum at $x = e$. Moreover, since the exponential function is monotonically increasing, the function $z = e^{x/\ln x}$ has a global minimum at $x = e$ as well. Thus, $z \geq e^e$, with equality at the point $x = y = e$.