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.
 A: If $x\leq 0$, $\sin x-x\geq 0$, so we just need to prove the inequality for $x>0$. Since in such a case $\sin x>-x$, we only need to prove the inequality for $x\in(0,6]$. Since $\sin x\geq-1$, we only need to prove it on $(0,4]$. Let $f(x)=x^2+3\sin x-3x$. Since $f(\pi)>0$ and $f'(x)>0$ on $[\pi,4]$, we only need to prove the inequality on $I=(0,\pi)$. Over such an interval, we can use the inequality:
$$ \sin(x)\geq \frac{1}{\pi}x(\pi-x) \tag{1}$$
and check that the second-degree polynomial given by replacing $\sin x$ with the RHS of $(1)$ is non-negative over $I$.
A: 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 $2\pi-6>0$.
A: A set of thoughts: Turning points are where $\cos x = 1 - (2/3)x$. One solution is when $x=0$ where the function has a local minimum of $0$. All solutions have $\sin x = \sqrt (9-(3-2x)^2)/3$ at which points the function has value $2\sqrt (3x-x^2) + (3x-x^2)$ which we can argue is non negative for real solutions. Hence all turning points are at non negative function values and since at $x=0$ the function has a local minimum the function is non negative for all $x$.
A: Let $f(x)=x^2+3\sin x-3x$.
When $x=0$, equality holds.
When $x<0$, let $y=-x$. Then $f(x)=y^2+3(y-\sin y)>0$.
When $x>4$, we have $x^2+3\sin x-3x> 4x-3-3x=x-3>0$.
When $x\in[\pi,4]$, $f(x)>0$ because $f(\pi)=\pi^2-3\pi>0$ and
$$f'(x)=2x+3\cos x-3=2(x-3)+3(1+\cos x)>0.$$
When $x\in I=(0,\frac\pi2)$, the equation $f''(x)=2-3\sin x=0$ has only one root $x_0=\arcsin\frac23$. Since
$$
f'(x_0)=2\arcsin\frac23+\sqrt{5}-3
\ >\ 2\left(\frac23\right)+\sqrt{5}-3
\ >\ 0,
$$
and at the endpoints of the closure of $I$, we have $f'(0)=0$ and $f'(\frac\pi2)=\pi-3>0$, we infer that $f'$ is positive on $I$. Yet $f(0)=0$. Therefore $f$ is also positive on $I$.
When $x\in(\frac\pi2,\pi)$, since $f'(x)$ is positive on $I$, we have
$$
f(x)=\int_0^x f'(t)dt>\int_{\pi-x}^x f'(t)dt=\int_{\pi-x}^{\frac\pi2} \left(f'(t)+f'(\pi-t)\right)dt=\int_{\pi-x}^{\frac\pi2} (2\pi-6)dt>0.
$$
