Prove that \(x - \frac{x^3}{3} \le \arctan x \le x\) for all \(x \ge 0\).
Solution
For every \(t\ge0\), \[ 1-t^2\le \frac1{1+t^2}\le1. \] The right inequality is immediate. For the left, \[ \frac1{1+t^2}-(1-t^2) =\frac{t^4}{1+t^2}\ge0. \] Integrating from \(0\) to \(x\ge0\) yields \[ \int_0^x(1-t^2)\,dt \le\int_0^x\frac{dt}{1+t^2} \le\int_0^x1\,dt. \] Since the middle integral is \(\arctan x\), this becomes \[ \boxed{x-\frac{x^3}{3}\le\arctan x\le x}. \]