<s>[INST] $a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ If $a$ and $b$ are positive rational numbers, prove that
$$a^a\cdot b^b\ge \left(\frac{a+b}{2}\right)^{a+b} \ge a^b \cdot{b^a}$$
My try:
consider $\frac{a}{b}$ and $\frac{b}{a}$ be two positive numbers with associated weights $b$ and $a$.
Then $\displaystyle\frac{b\cdot\frac{a}{b}+a\cdot\frac{b}{a}}{a+b}\ge \biggl[\left(\frac{a}{b}\right)^b\cdot \left(\frac{b}{a} \right)^a\biggl]^\frac{1}{a+b}$ implies $a^a\cdot{b^b}\ge a^b\cdot{b^a}$
Please help me to solve this problem. Thanks
 [/INST] Part 1: by GM-HM:
$\left(\underbrace{(a.a.a\dots a)}_{\text{a times}}\underbrace{(b.b.b\dots b)}_{\text{b times}}\right)^{\frac{1}{a+b}} \ge \frac{a + b}{\left(\underbrace{\frac{1}{a}+ .... \frac{1}{a}}_{\text{a times}}\right)+\left(\underbrace{\frac{1}{b}+ .... \frac{1}{b}}_{\text{b times}}\right)}$
$\implies a^ab^b \ge \left(\frac{a+b}{2}\right)^{a+b}$
Part 2: by AM-GM:
$\frac{\underbrace{(a+a+a+\dots +a)}_{\text{b times}}+\underbrace{(b+b+b\dots +b)}_{\text{a times}}}{a+b} \ge \left(a^bb^a\right)^{\frac{1}{a+b}}$
$\implies \frac{2ab}{a+b} \ge \left(a^bb^a\right)^{\frac{1}{a+b}}$
$\implies \frac{a+b}{2} \ge \frac{2ab}{a+b} \ge \left(a^bb^a\right)^{\frac{1}{a+b}}$ (using AM-HM for the first inequality)
$\implies \left(\frac{a+b}{2}\right)^{a+b} \ge a^bb^a $
</s>