Prove that $$x\cos x \le \frac{\pi^2}{16}$$for $0 \le x \le \frac{\pi}{2}$
Problem
Source: All-Russian MO 2009 Regional 11. 3
Tags: algebra, inequalities, trigonometry
mavanasavana
28.08.2024 21:18
Consider the function $f(x)=\frac{{{\pi }^{2}}}{16}-x\cos x$ and because $f'(x)=\cos x\left( x-\tan x \right)\ge 0,\forall x\in \left[ 0,\frac{\pi }{2} \right]\Rightarrow f$ increasing, hence $f(x)\ge f(0)=0$ and done!
arqady
28.08.2024 22:36
mavanasavana wrote:
Consider the function $f(x)=\frac{{{\pi }^{2}}}{16}-x\cos x$ and because $f'(x)=\cos x\left( x-\tan x \right)\ge 0,\forall x\in \left[ 0,\frac{\pi }{2} \right]\Rightarrow f$ increasing, hence $f(x)\ge f(0)=0$ and done!
I think, $f'(x)=-\cos{x}+x\sin{x}=\cos{x}(x\tan{x}-1).$ Also, I think, $f(0)\neq0.$
rms
28.08.2024 23:12
parmenides51 wrote: Prove that $$x\cos x \le \frac{\pi^2}{16}$$for $0 \le x \le \frac{\pi}{2}$ For $0 \le x \le \frac{\pi}{2}$ function $f(x)=x+\cos x -\frac{\pi}{2}$ is increasing and $f({\pi}/2)=0$ so $x+\cos x \le \frac{\pi}{2}$. Then $x\cos x \le \frac14(x+\cos x)^2\le \frac{\pi^2}{16}$.