Is there a shorter or more trivial way to prove that $ x > \cos (x)-\cos (2 x) $ holds for all $x>0$? I want to prove that the inequality
$$ x > \cos (x)-\cos (2 x)  $$
holds for all $x>0$.
My attempt:
Since the function on the RHS is periodic, we can find the position of extrema (on first period $2\pi$) by computing the first derivative; comparing them with the function $x$ which is increasing, we can prove that it holds.

Question
Is there a shorter or more trivial/obvious way to prove this?
 A: This can be seen geometrically.
Let $P = (\cos x, \sin x)$ and $Q = (\cos 2x, \sin 2x)$; then arc $PQ$ on the unit circle has length $x$. Its projection onto the horizontal axis is an interval containing both $\cos x$ and $\cos 2x$, and the length of that projection is always smaller, because arc $PQ$ is not perfectly horizontal and flat. Therefore
$$
    x > |{\cos x - \cos 2x}|
$$
for all $x>0$, which implies the inequality we want.
A: If $x>2$, the statement is trivial. For $0<x\leq 2,$ one has $$x>2\sin\left(\frac x2\right)\geq 2\sin\left(\frac{3x}2\right)\sin\left(\frac x2\right)=\cos(x)-\cos(2x).$$
A: Almost solution.
Mean value theorem.  Note $\cos' = -\sin$.  For $x>0$ we have $ 0 < x < 2x$ and
$$
\frac{\cos(2x) - \cos(x)}{2x-x} = -\sin(\xi)
$$
for some $\xi$ between $x$ and $2x$.  Thus
$$
\frac{\cos(2x) - \cos(x)}{x} \le 1
\quad\text{so}\quad
\cos(2x)-\cos(x) \le x .
$$
Why can we not have equality?  For example, $\xi = 3\pi/2$?
A: For $x > 0$ is
$$
 \cos(x)-\cos(2x) = \int_x^{2x} \sin(t) \, dt < \int_x^{2x} 1 \, dt  = x \, .
$$
Strict inequality between the integrals holds because $\sin(t) < 1$ for “almost all” $t$.
