If $a+b=1$ where $a,b\in\Bbb R^+$ then prove that $a^ab^b+b^aa^b\le1$

Question

If $a+b=1$ such that $a$ and $b$ are positive real numbers, then prove $$a^a b^b+b^a a^b\le1$$

I tried applying Arithmetic and Geometric mean but it doesn't work. I also tried to equate the given equation with $a^0$( which would be equal to $1$), but, as it turns out, that doesn't work either

Answer 1

By AM-GM $a^ab^b+a^bb^a\leq a\cdot a+b\cdot b+b\cdot a+a\cdot b=1$.