TITLE: Proof that $x^y + y^x > 1 \ \forall x,y > 0 $
QUESTION [8 upvotes]: The problem is trivial if at least one of $x$ or $y$ is greater than $1$.  So all we need is to proof that $x^y+y^x > 1 \ \forall x,y \in (0,1)$.

REPLY [2 votes]: By Bernoulli 
$$\frac{1}{x^y}=\left(1+\frac{1}{x}-1\right)^y\leq1+y\left(\frac{1}{x}-1\right)=\frac{x+y-xy}{x}.$$
Thus,
$$x^y\geq\frac{x}{x+y-xy}>\frac{x}{x+y}$$
and we are done.