<s>[INST] Proof of inequality $x^2+3\sin x-3x \geq 0$ I'm looking for a simple (or any) way to prove the following inequality:
$$x^2+3\sin x-3x \geq 0\quad \text{for all }x\in \mathbb R$$
Progress
I have looked at the function corresponding to the left hand side: it has a root at zero, but its derivatives do not seem to give me useful information.
 [/INST] For $x\le 0$ we have $\sin x\ge x$ and hence $$x^2+ 3\sin x-3x\ge x^2\ge 0.$$ If $x\ge 4$ we have
$$ x^2+ 3\sin x-3x\ge x^2-3x-3=(x-4)(x+1)+1>0.$$
Hence we need only consider $0<x<4$.
Taylor says $$\begin{align}f(x)
&=f(a)+f'(a)(x-a)+\frac12f''(a)(x-a)^2+\frac16f'''(\eta)(x-a)^3\\\end{align}$$
with $\eta$ between $a$ and $x$. Here, with $f(x)=3\sin x-3x+x^2$ and $a=0$ (so that $f'(x)=3\cos x-3+2x$, $f''(x)=-3\sin x+2$, $f''(x)=-3\cos x$)
$$ \tag1f(x) =x^2\cdot\left(1-\frac x2\cos\eta\right)$$
and with $a=\pi$
$$ \tag2f(x) =\underbrace{(\pi-3)\pi +(2\pi-6)(x-\pi)}_{=:g(x)}+(x-\pi)^2\left(1-\frac{x-\pi}2 \cos\eta\right)$$
The right hand side in $(1)$ is positive for $0<x<2$. The last summand in  $(2)$ is nonnegative for $\pi-2\le x\le \pi+2$, so especially for $2\le x<4$.
Hence to complete the proof ist suffices to show $g(x)>0$ for $x\ge 2$. This follows from  $g(\pi/2)=0$ together with $\frac\pi2<2$ and $2\pi-6>0$.
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