2014

Abstract

For $\pi /2\le x\le \pi ,$ it holds $\cos x\le 0<e^{-x^2/2},$ hence it remains to consider $0<x<\pi /2$.After taking the logarithm on both sides the inequality turns to $\ln \,\cos x<-x^2/2,$ or equivalently ${x^2/2+\ln \,\cos x<0},$ $0<x<\pi /2$. The function $f(x)=x^2/2+\ln \,\cos x$ is decreasing on $[0,\pi /2),$ because $f'(x)=x-\tan x<0,$ ${0<x<\pi /2.}$ Therefore, $f(x)<f(0)=0,$ $0<x<\pi /2$.