Prove that $x^y < y^x$
Hint:
This is equivalent to $y\ln x<x\ln y$, i.e. since $x, y>0$, to $$\frac{\ln x}x<\frac{\ln y}y.$$
Set $f(x)=\dfrac{\ln x}x$ and determine the variations of $f$.
HINT
take logs to compare $x \ln y ? y \ln x$ or equivalently $\frac{x}{\ln x} ? \frac{y}{\ln y}$ (sign does not change since $\ln x > \ln y > 1$) and look at the function $x/\ln x$ to see if it's increasing or decreasing for $x > e$