<s>[INST] Proof of $\sin2x+x\sin^2x \lt\dfrac{1}{4}x^2+2$ How can be proven the following inequality?
$$\forall{x\in\mathbb{R}},\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$$
Thanks
 [/INST] For $x<0$ inequality is obvious.
$\displaystyle(\frac{x}{2}-1)^2=\frac{x^2}{4}-x+1 \geq 0$, so
$\displaystyle \frac{x^2}{4}+2 \geq x+1$, but for $x \geq 0$
$x \geq x\ (\sin x)^2$ and $1 \geq \sin 2x$
If you want to have $\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$ instead of $\left[\sin(2x)+x\sin(x)^2\right]\le\dfrac{1}{4}x^2+2$ you must have $\sin x=\pm 1$ and $\sin 2x=1$, but it's impossible.
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