Consider function $f(x)=x^2 - (1-\cos{x})$. It is clear that $f(0)=0$. We are going to show $f(x)$ is an increasing function for $x > 0$. If so, that $f(x) > 0$ for all $x > 0$ which leads to the claim immediately.

We have $f'(x) = 2x - \sin{x}$ and $f''(x)=2-\cos{x} > 0$. Hence, $f'(x)$ is strictly increasing. Meanwhile, we have $f'(0) = 0$, Therefore $f'(x) > 0$ for all $x > 0$. The last statement implies $f(x)$ strictly increases for all $x > 0$.