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.
 A: The extrema of the function correspond to the roots of the first derivative.
$$f'(x)=2x-2\sin(x)=0.$$
You only have the solution $x=0$ because for other values $|\sin(x)|<|x|$.
As the function goes to $+\infty$, the extremum must be a minimum, and it has the value
$$f(0)=0+2\cos(0)-2=0.$$

If you don't want to take derivatives, start from
$$\left|\sin\left(\frac x2\right)\right|\le\left|\frac x2\right|$$ or
$$4\sin^2\left(\frac x2\right)\le x^2$$
By the doubling formula ($\cos(2t)=1-2\sin^2(t)$), this becomes
$$0\le x^2-4\sin^2\left(\frac x2\right)=x^2+2\cos(x)-2.$$

The relation $|\sin(x)|\le|x|$ can be verified from geometric considerations. On the trigonometric circle, $2x$ is the length of an arc of aperture angle $2x$, while $2\sin(x)$ is the length of the chord for the same angle. The chord is the shortest distance.

A: 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 $x-\sin(x) \ge 0$
A: 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 $2$…
