Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to determine the critical points. So:
$\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $
The critical points:
$\cos(x) -1 + \frac{x^2}{2} =  0 \\ $
It seems that x = 0 is a critical point. 
Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' =  -\sin(x) + x \\ $
and $-\sin(0) + 0 = 0 \\$
The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0).
Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?

You just have to prove your inequality when $x\in(0,\pi)$, since otherwise the RHS is below $-1$. Consider that for any $x\in(0,\pi/2)$,
$$ \sin^2 x < x^2 \tag{1}$$
by the concavity of the sine function. By setting $x=y/2$, $(1)$ gives:
$$ \forall y\in(0,\pi),\qquad \frac{1-\cos y}{2}<\frac{y^2}{4}\tag{2}, $$
so:
$$ \cos y > 1-\frac{y^2}{2} \tag{3} $$
for any $y\in(0,\pi)$. By integrating $(3)$ with respect to $y$ over $(0,x)$ we get our inequality.
