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

Question

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

Answer 1

The second inequality.

We need to prove that: $$\ln\frac{a+b}{2}\geq\frac{b}{a+b}\ln{a}+\frac{a}{a+b}\ln{b}.$$ Now, since $\frac{b}{a+b}+\frac{a}{a+b}=1$ and $\ln$ is a concave function, by Jensen and AM-GM we obtain: $$\frac{b}{a+b}\ln{a}+\frac{a}{a+b}\ln{b}\leq\ln\left(\frac{ba}{a+b}+\frac{ab}{a+b}\right)\leq\ln\frac{a+b}{2}.$$

Answer 2

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 $$\left(\frac{a+b}{2}\right)^{a+b} \le a^ab^b$$

Answer 3

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.

Answer 4

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 $