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.

Given that both $a$ and $b$ are positive integers, let us consider the case where $b > a$.
$b$ can be expressed as $a+x$, where $x$ is some positive integer.
to prove $a^a b^b > a^b   b^a$,we need to prove that $a^a   b^b - a^b  b^a > 0$
Rewrite $a^a   b^b - a^b   b^a$, by substituting $b = (a+x)$
$= a^a (a+x)^{a+x} - a^{a+x}  (a+x)^a$
expanding the powers
$= a^a  (a+x)^a  (a+x)^x - a^a  a^x  (a+x)^a$
taking common factors out of bracket, we get 
$= a^a  (a+x)^a  [ (a+x)^x - a^x ]  $
since $x \gt0$, $(a+x)^x$ must be greater than $a^x$, therefore the above expression evaluates to a value greater than zero.
The other case of $a\gt b$ is the same as this since the order of $a$ and $b$ don't matter, while $a=b$ is trivial.
