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, \(f(x)<f(0)=0,\) \(0<x<\pi /2\).
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Brayman, V., Kukush, A. (2017). 2014. In: Undergraduate Mathematics Competitions (1995–2016). Problem Books in Mathematics. Springer, Cham. https://doi.org/10.1007/978-3-319-58673-1_42
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DOI: https://doi.org/10.1007/978-3-319-58673-1_42
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