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.

We'll show a bit more: $a^a b^b > \left( \frac{a+b}{2} \right)^{a+b} > a^b b^a.$
First we need a lemma: $P = (1+x)^{1+x} (1-x)^{1-x} > 1$ if $x < 1$.
Proof: $\ln P = (1+x)\ln (1+x) + (1-x) \ln (1-x) = x ( \ln (1+x) - \ln (1-x)) + \ln (1+x) + \ln (1-x) = 2x \left(x + \frac{x^3}{3}+\frac{x^5}{5} + \cdots \right) - 2 \left( \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \cdots \right) = 2 \left(\frac{x^2}{1 \cdot 2} + \frac{x^4}{3 \cdot 4} + \frac{x^6}{5 \cdot 6} + \cdots \right).$
Therefore, $\ln P$ is positive, and so $P > 1.$
In this lemma put $x = \frac{s}{t}$, where $t > s$.  Then
$ \left( 1 + \frac{s}{t} \right)^{1+ \frac{s}{t}} \left(1 - \frac{s}{t}\right)^{1-\frac{s}{t}} > 1. $  Thus
$ \left( \frac{t+s}{t} \right)^{t+s} \left(\frac{t-s}{t} \right)^{t-s} > 1^t = 1.$
Thus $ (t + s)^{t+s}(t - s)^{t-s} > t^{2t}$.  Letting $t + s = a$ and $t - s = b$, we see that $t = \frac{a+b}{2}$ and so $a^a b^b > \left( \frac{a+b}{2}\right)^{a+b}.$
In a similar way, letting $Q = (1+x)^{1-x} (1-x)^{1+x}$, we can show that $\ln Q < 0$ and so $Q < 1.$  Now continue as we did in the last proof.
