If both $a,b>0$, then $a^ab^b \ge a^bb^a$ Prove that $a^a \ b^b \ge a^b \ b^a$, if both $a$ and $b$ are positive.

$$a^a \ b^b \;?\; a^b \ b^a \\
\frac{a^a}{b^a} \;?\; \frac{a^b}{b^b} \\
\left(\frac{a}{b}\right)^a \;?\; \left(\frac{a}{b}\right)^b \\
\left(\frac{a}{b}\right)^{a-b} \;?\; 1 $$
if $a \ge b$, then $c = \frac{a}{b} \ge 1$, and $d = a-b \ge 0$. Thus $c^d \ge 1$, so $?$ is $\ge$.
if $a < b$, then:
$$\left(\frac{a}{b}\right)^{a-b} \\
= \left(\frac{b}{a}\right)^{b-a}$$
and we re-use the first result by symmetry.
