By the Mean value theorem show that:
$\displaystyle \arctan{x} \leq x\ for\ x \geq 0$
Assume that
$\displaystyle \arctan{x} > x$ for $\displaystyle x \geq 0$.
Then by differentiating both sides
$\displaystyle \frac{1}{1 + x^2} > 1$
$\displaystyle 1 > 1 + x^2$
$\displaystyle x^2 < 0$.
But $\displaystyle x^2 \geq 0$ for all $\displaystyle x \in \mathbf{R}$. So we have arrived at a contradiction. So our original statement that $\displaystyle \arctan{x} > x$ is wrong.
Therefore $\displaystyle \arctan{x} \leq x$ for all $\displaystyle x \geq 0$.