Proof of $\sin2x+x\sin^2x \lt\dfrac{1}{4}x^2+2$

Question

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

Answer 1

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.

Answer 2

Hint:

$\sin(2x)+x\sin^2x \leq 1+x$ and $(x-2)^2\geq0$

Answer 3

$$\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$$ $$\sin(x)(2\cos(x)+x\sin(x))\lt\dfrac{1}{4}x^2+2$$ while the right side is positive for all $x \in \mathbb{R}$ it is enoght that we show: $$|\sin(x)|\cdot|(2\cos(x)+x\sin(x))|\lt\dfrac{1}{4}x^2+2$$ by the Cauchy–Schwarz inequality we have: $$|\sin(x)|\cdot\sqrt{|4+x^2|.|\sin(x)^2+\cos(x)^2|}\lt\dfrac{x^2+4}{4}+1$$ $$|\sin(x)|\lt\dfrac{\sqrt{|4+x^2|}}{4}+\dfrac{1}{\sqrt{|4+x^2|}}$$ and the second side is greater than 1 by arithmetic and geometric inequality and equality.