If $ab>0$ and $\displaystyle 0<x<\frac{\pi}{2}$, prove that \[ \left ( 1+\frac{a^2}{\sin x} \right ) \left ( 1+\frac{b^2}{\cos x} \right ) \geq \frac{(1+\sqrt{2}ab)^2 \sin 2x}{2}. \]
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Tags: trigonometry, inequalities, inequalities unsolved
djmathman
14.10.2014 05:10
Note that from Cauchy-Schwarz we obtain \[(2\sin x+2a^2)(2\cos x+2b^2)\geq(\sqrt{4\sin x\cos x}+2ab)^2=(\sqrt{2\sin 2x}+2ab)^2.\] Now remark that from $1\geq\sin 2x$ we obtain \[\sqrt{2\sin 2x}+2ab\geq \sqrt{2\sin 2x}+ab\sqrt{4\sin 2x}=\sqrt{2\sin 2x}(1+\sqrt2ab).\] Thus squaring gives \begin{align*}4(\sin x+a^2)(\cos x+b^2)\geq(\sqrt{2\sin 2x}+2ab)^2&\geq 2\sin 2x(1+\sqrt2ab)^2\geq \sin^2 2x(1+\sqrt2ab)^2\\\implies \left(1+\dfrac{a^2}{\sin x}\right)\left(1+\dfrac{b^2}{\cos x}\right)&\geq\dfrac{(1+\sqrt2ab)^2\sin 2x}{2},\end{align*} where we divided $4\sin x\cos x$ from the first inequality to get the second.
Kunihiko_Chikaya
14.10.2014 10:04
Since $0<2x<\pi$, we have $\sin x\cos x=\frac 12 \sin 2x\leq \frac 12,\ \sin x,\ \cos x >0$ and $ab>0$, by Cauchy-Schwarz inequality, $\left(1+\frac{a^2}{\sin x}\right)\left(1+\frac{b^2}{\cos x}\right)\geq \left(1+\frac{ab}{\sqrt{\sin x\cos x}}\right)^2$ $\geq (1+\sqrt{2}ab)^2)>\frac{(1+\sqrt{2}ab)^2}{2}\sin 2x\ (\because 0<\sin 2x<1)$ Equality holds when $\sin 2x=1\ \left(0<x<\frac{\pi}{2}\right)$ and $\left(1,\ \frac{a^2}{\sin x}\right)\parallel \left(1,\ \frac{b^2}{\cos x}\right)$ $\Longleftrightarrow x=\frac{\pi}{4}, \ \tan x=\frac{a}{b}.$