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.

It is enough to prove it for $x\ge 0$. Now, if $x\ge 0$, $(x^2+2\cos x)'=2(x-\sin x)\ge 0$, hence $x^2+2\cos x$ is non-decreasing for $x\ge 0$, and its value for $x=0$ is $2$…
