Inequality $\arctan x ≥ x-x^3/3$ Can you help me prove $\arctan x ≥ x-x^3/3$? I have thought of taylor but I have not come up with a solution.

Have you tried derivatives?
$$(\arctan x-x+x^3/3)'=\frac 1{1+x^2}-1+x^2=\frac{x^4}{1+x^2}\ge0$$
So the difference is an increasing function. This fact, together the equality when $x=0$ means that
$$\arctan x\ge x-x^3/3\text{ when }x\ge 0\\\arctan x\le x-x^3/3\text{ when }x\le 0$$
