Prove that x – (1/3)x³ < arctan x if x is positive

by RoRi

November 29, 2015

Consider the function

$f(x) = \arctan x - x + \frac{x^3}{3}.$

Compute the derivative $f'(x)$ and use the sign of the derivative to prove the inequality

$x - \frac{x^3}{3} < \arctan x \qquad \text{for all } x > 0.$

Proof. First, we compute the derivative

$\begin{align*} f'(x) &= \frac{1}{1+x^2} - 1 + x^2 \\ &= \frac{x^4}{1+x^2}. \end{align*}$

Therefore, $f'(x) > 0$ for all $x \neq 0$ . This implies $f(x)$ is strictly increasing if $x \neq 0$ , and in particular, $f(x)$ is increasing for $x > 0$ . This implies

$\begin{align*} f(x) > f(0) && \implies && \arctan x - x + \frac{x^3}{3} &> 0 \\ && \implies && x - \frac{x^3}{3} &< \arctan x. \qquad \blacksquare \end{align*}$

Stumbling Robot

Prove that x – (1/3)x³ < arctan x if x is positive

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