Prove that for all $a, b > 0, a^b + b^a > 1$

Question

This seems an innocent inequality.
The trivial case when $a \ge 1$ or $b \ge 1$ is easy to prove:
if $a \ge 1$, then $a^b \ge a^0 = 1$.
So the hard part of the problem is when $a, b \in (0, 1)$ .