Question

Prove that ${\log }_{\pi }2+{\log }_2\pi \ge 2$

Answer 1

Arithmetic Mean of ${\log }_{\pi }2,{\log }_2\pi$ is $\frac{{\log }_{\pi }2+{\log }_2\pi }{2}$

Geometric Mean of ${\log }_{\pi }2,{\log }_2\pi$ is $\sqrt{{\log }_{\pi }2{\log }_2\pi }$

Since, $A.M\ge G.M.$
$\frac{{\log }_{\pi }2+{\log }_2\pi }{2}\ge \sqrt{{\log }_{\pi }2{\log }_2\pi }=1$ ($\because {\log }_{a}b=\frac{1}{{\log }_{b}a}$)

$\frac{{\log }_{\pi }2+{\log }_2\pi }{2}\ge 1$

$\Rightarrow {\log }_{\pi }2+{\log }_2\pi \ge 2$