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.

$$
1 - t^4 \le 1
$$
so 
$$
1-t^2 \le \frac1{1+t^2}
$$
now integrate 
$$
\int_0^x (1-t^2)dt \le \int_0^x \frac{dx}{1+t^2}
$$
i.e.
$$
x-\frac{x^3}3 \le \arctan x
$$
