Proof for $\sin(x) > x - \frac{x^3}{3!}$

A standard approach is to let $f(x)=\sin x-\left(x-\frac{x^3}{3!}\right)$, and to show that $f(x)\gt 0$ if $x\gt 0$.

Note that $f(0)=0$. We will be finished if we can show that $f(x)$ is increasing in the interval $(0,\infty)$.

Note that $f'(x)=\cos x-1+\frac{x^2}{2!}$. We will be finished if we can show that $f'(x)\gt 0$ in the interval $(0,\infty)$.

Note that $f'(0)=0$. We will be finished if we can show that $f'(x)$ is increasing in $(0,\infty)$.

So we will be finished if we can prove that $f''(x)\gt 0$ in the interval $(0,\infty)$.

We have $f''(x)=-\sin x+x$. Since $f''(0)=0$, we will be finished if we can show that $f'''(x)\ge 0$ on $(0,\infty)$, with equality only at isolated points. This is true.

Or else for the last step we can use the geometrically evident fact that $\frac{\sin x}{x}\lt 1$ if $x\gt 0$.

Remark: It is more attractive to integrate than to differentiate, but we used the above approach because differentiation comes before integration in most calculus courses.

For the integration approach, let $x$ be positive. Since $\sin t\lt t$ on $(0,x)$, we have $\int_0^x (t-\sin t)\,dt\gt 0$. Integrate. We get $\cos x+\frac{x^2}{2}-1\gt 0$ (Mean Value Theorem for integrals), so $\cos t+\frac{t^2}{2}-1\gt 0$ if $t\gt 0$.

Integrate from $0$ to $x$. We get $\sin x+\frac{x^3}{3!}-x\gt 0$, or equivalently $\sin x\gt x-\frac{x^3}{3!}$. Nicer, by a lot.

Take a decreasing sequence of positive real numbers $a_n$ such that $a_n\to 0$.

Now, consider the sequence $b_k=\sum_{n=1}^k (-1)^{n-1}a_n$. The alternating series criterion guarantee us that it converges to some $b$.

Note that $b_1=a_1$, $b_2=b_1-a_2\in(0,b_1)$, $b_3=b_2+a_3\in(b_2,b_1)$, etc. So the limit $b$ is lesser that the terms $b_{2k+1}$ and greater than $b_{2k}$.

Then, if $x<\sqrt 6$, $$\sin x=\sum_{n=1}^\infty(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}>\sum_{n=1}^2(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}=x-\frac{x^3}{3!}$$

If $x\geq\sqrt 6$, the function $f(x)=x-x^3/6$ is decreasing and $f(\sqrt 6)=0$, so $f(x)<0$ for $x>\sqrt 6$. Since $\sin x>0$ for $0<x<\pi$, we have that $\sin x>f(x)$ for $0<x<\pi$. (Note that $\sqrt 6<\pi$).

Last, for $x\geq \pi$, $\sin x\geq -1$ and $f(x)<f(\pi)<f(3)=3-4.5<-1$.

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.