IIT JEE , AIEEE Forum - Mathematics , Algebra

ramkumar_november

$f(x)\;=\;(tan^{-1}x)^2\;+\;\frac{2}{\sqrt{x^2+1}}\;\;\;x>0$

$let\;\;g(x)\;=\;tan^{-1}x\;-\;\frac{x}{\sqrt{x^2+1}}$

$therefore\;\;f$

$f(x)\;is\;an\;increasing\;function\;for\;all\;x>0$

$\frac{1}{e}\;<\;e$

$therefore\;f(\frac{1}{e})\;<\;f(e)$

$thus\;we\;get\;\;(tan^{-1}\bigg(\frac{1}{e}\bigg))^2\;\;+\;\;\frac{2e}{\sqrt{e^2+1}}\;\;<\;\;(tan^{-1}e)^2\;+\;\frac{2}{\sqrt{e^2+1}}$