Problem: Prove that for all $x>0$ we have \[ \sin x > x-\frac{x^3}{6}. \]
Solution: Let \[ f(x) = \sin x-x+\frac{x^3}{6}. \] We will show that the function is strictly increasing for $x>0$. As $f(0)=0$ so for every $x>0$, the function $f(x)>0,$ which will prove the inequality.
Consider the derivatives of $f$. \begin{align*} & f^\prime (x) = \cos x-1+\frac{x^2}{2},\quad f^\prime (0)=0 \\ & f^{\prime\prime}(x) = -\sin x + x, \quad f^{\prime\prime }(0) = 0\\ & f^{\prime \prime \prime }(x) = -\cos x+1. \end{align*} Now note that \[ f^{\prime \prime \prime }(x) \ge 0,~\forall~x\in \mathbb{R} . \] The equality will be achieved on the discrete set $\{2\pi n:n\in \mathbb{Z} \}$.
The following result is easy to prove. The main idea is that the property of increasing is a local property.
Results: If for a function $f$ the derivative $f^\prime (x)>0$ except for a discrete set, then $f$ is strictly increasing function.
Therefore, the function $f^{\prime \prime }$ is strictly increasing function on $\mathbb{R} $. Now note that $f^{\prime \prime }(0)=0$, so for every $x>0$, we have $f^{\prime \prime }(x)>0$, so $f^\prime $ is strictly increasing function on $\mathbb{R} ^+$. Again, $f^\prime (0)=0$ so $f^\prime (x)>0$ for every positive $x$. Thus, $f$ is a strictly increasing function on $\mathbb{R} ^{+}$. Hence, \[ \textcolor{blue}{\boxed{\sin x > x-\frac{x^3}{6},\quad \forall~x>0}} \]