MHB Maximize 2sinxcosx/[(1+sinx)(1+cosx)]

[anemone]

Maximize $\dfrac{2\sin x \cos x}{(1+\sin x)(1+\cos x)}$ for $x\in \left(0, \dfrac{\pi}{2}\right)$.

 

[anemone]

Spoiler: Hint

AM-GM inequality

 

[anemone]

Spoiler: My solution

$\begin{align*}\dfrac{2\sin x \cos x}{(1+\sin x)(1+\cos x)}&=\dfrac{2\sin x \cos x (1-\sin x)}{(1-\sin^2 x)(1+2\cos^2 \dfrac{x}{2}-1)}\\&=\dfrac{2\sin x \cos x (1-\sin x)}{(\cos^2 x)(2\cos^2 \dfrac{x}{2})}\\&=\dfrac{\sin x}{\cos x}\left(\dfrac{1-2\sin \dfrac{x}{2} \cos \dfrac{x}{2}}{\cos^2 \dfrac{x}{2}}\right)\\&=\tan x\left(\sec^2 \dfrac{x}{2}-2\tan \dfrac{x}{2}\right)\\&=\dfrac{2\tan \dfrac{x}{2}}{1-\tan^2 \dfrac{x}{2}}\left(1+\tan^2 \dfrac{x}{2}-2\tan \dfrac{x}{2}\right)\\&=2\tan \dfrac{x}{2}\dfrac{\left(1-\tan \dfrac{x}{2}\right)\left(1-\tan \dfrac{x}{2}\right)}{\left(1+\tan \dfrac{x}{2}\right)\left(1-\tan \dfrac{x}{2}\right)}\\&=2\tan \dfrac{x}{2}\left(\dfrac{\tan \dfrac{\pi}{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi}{2}\tan \dfrac{x}{2}}\right)\\&=2\tan \dfrac{x}{2}\tan \left( \dfrac{\pi}{4}-\dfrac{x}{2}\right)\end{align*}$

Therefore by the AM-GM inequality,

$\begin{align*}\sqrt{\dfrac{2\sin x \cos x}{(1+\sin x)(1+\cos x)}}&=\sqrt{2\tan \dfrac{x}{2}\tan \left( \dfrac{\pi}{4}-\dfrac{x}{2}\right)}\le \dfrac{\sqrt{2}}{2}\left(\tan \dfrac{\pi}{2}+\tan \left( \dfrac{\pi}{4}-\dfrac{x}{2}\right)\right)\end{align*}$

Equality attains when $\tan \dfrac{\pi}{2}=\tan \left( \dfrac{\pi}{4}-\dfrac{x}{2}\right)$, i.e. when $x=\dfrac{\pi}{4}$ where $\tan \dfrac{\pi}{8}=\sqrt{2}-1$, an exact value we could get from using the double angle formula for $\tan x$ and that $\tan \dfrac{\pi}{4}=1$.

Hence $\dfrac{2\sin x \cos x}{(1+\sin x)(1+\cos x)}\le 2(\sqrt{2}-1)^2=2(3-2\sqrt{2})$.