$a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ 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

For the first inequality:
Let $f(x) = x\ln x$, then $f''(x) = 1/x > 0$ and hence convex. By the definition of a convex function, we have
$$f\left(\frac{a+b}{2}\right) \le \frac{f(a)+f(b)}{2}$$
which, in our case, becomes
$$\frac{a+b}{2} \ln \left(\frac{a+b}{2}\right) \le \frac{a\ln a + b\ln b}{2}$$
or equivalently
$$\left(\frac{a+b}{2}\right)^{a+b} \le a^ab^b$$
