<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] From weighted AM-GM, we have $$\left(\frac{\frac 1a\cdot a+\frac 1 b\cdot b}{a+b}\right )^{a+b}\geq \frac{1}{a^ab^b}\\ \implies \left(\frac{2}{a+b}\right)^{a+b}\geq \frac{1}{a^ab^b}$$which is equivalent to the first inequality.
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