Show $x^2 + 2\cos(x) \geq 2$ How do I show that $x^2 + 2\cos(x) - 2$ is always nonnegative (x is measured in radians)? If $x \geq 2$ or $x \leq -2$ then obviously, $x^2 \geq 4$, and so it must be true. But otherwise, $2\cos(x)$ can be as small as $-2$ and it is quite surprising that something that could potentially be as small as $x^2 - 4$ is never actually negative. I'm not sure how to go about solving this, especially since there is an $x^2$ term which is annoying.
Edit: Preferably without calculus, although the existing calculus answers are fine.

Since the functions are even, we only consider the region [0,2].
For $x= 0$,  we have $2\ge 2$ is true.
For $x > 0$, we show the derivative of the function is positive. That is the function is increasing.
$2x - 2 \sin(x) = 2(x-\sin(x)) \ge 0$ as $x-\sin(x) \ge 0$
