Prove that $\log_a\left(\frac{2ab}{a+b}\right)\log_b\left(\frac{2ab}{a+b}\right)\geq 1$

Question

Prove that $$\log_a\left(\frac{2ab}{a+b}\right)\log_b\left(\frac{2ab}{a+b}\right)\geq 1$$ if a and b are between 0 and 1 I'm stuck.It should be solved using one of the four means or by getting an obvious statement. Not sure which one and how to solve it.

Answer 1

For $0<a,b<1$ let's assume $c=\frac1a$ $d=\frac1b$

$$\log_a\left(\frac{2ab}{a+b}\right)\log_b\left(\frac{2ab}{a+b}\right)\geq 1 \iff$$

$$\frac{\log\left(\frac{2ab}{a+b}\right)\log\left(\frac{2ab}{a+b}\right)}{\log a \log b}\geq 1 \iff$$

$$\frac{\log\left(\frac{c+d}{2}\right)\log\left(\frac{c+d}{2}\right)}{\log c \log d}\geq 1 \iff$$

For AM-GM

$$\log \frac{c+d}{2}\geq \log \sqrt {cd} =\frac{\log c+\log d}{2} \geq \sqrt{\log c \log d}$$

Thus

$$\frac{\log\left(\frac{c+d}{2}\right)\log\left(\frac{c+d}{2}\right)}{\log c \log d}\geq \frac{\log c \log d}{\log c \log d}= 1 \quad \square$$

Answer 2

By AM-GM and C-S we obtain: $$\log_a\frac{2ab}{a+b}\log_b\frac{2ab}{a+b}\geq\log_a\frac{2ab}{2\sqrt{ab}}\log_b\frac{2ab}{2\sqrt{ab}}=\log_a\sqrt{ab}\log_b\sqrt{ab}=$$ $$=\frac{1}{4}\left(1+\log_ab\right)\left(1+\log_ba\right)\geq\frac{1}{4}\left(1+\sqrt{\log_ab\log_ba}\right)^2=1.$$