How to prove $\cos(x) \ge 1-\frac{2}{\pi}(x+\sin(x))$? I need help in solving the following inequality---
$$\cos(x) \ge 1-\frac{2}{\pi}(x+\sin(x))$$
The inequality is used in proving Margolus Levitin theorem as can be seen here.
I tried using the Cauchy Mean Value theorem, in similar ways as mentioned in Proving $\frac2\pi x \le \sin x \le x$ for $x\in [0,\frac {\pi} 2]$, and rearranging the inequality to a more manageable form, but couldn't proceed further.
Essentially the same question has been asked on Physics SE here
but the currently accepted answer is a bit heuristical because it assumes a specific form of expression for $\cos(x)$ with adjustable parameters and then finds the parameters by several arguments.
I am looking for a more formal and direct proof probably using some well-known inequalities (like Cauchy-Schwarz inequality) or theorems (like Mean Value Theorem).
EDIT: The domain should be $x>0$ because clearly the inequality doesn't hold for $x<0$ as pointed out in a comment by Greg Martin.
This is physically justifiable while using in the Margolus Levitin theorem because there $x\propto t$ and time $t>0$.
 A: This is equivalent to
$$
\frac{1-\cos x}{x+\sin x}\leq \frac{2}{\pi}
$$
for $x>0$. Take derivative of left we get $\frac{x\sin x}{(x+\sin x)^2}$, this is $\geq 0$ until $x=\pi$, when the left equals to $\frac{2}{\pi}$; after that the left decreases to $0$ until $x=2\pi$. After $2\pi$ the denominator is even bigger, e.g. the left $\leq \frac{1+1}{2\pi-1}<\frac{2}{\pi}$. Overall the maximum of left side over$[0, \infty)$ is $\frac{2}{\pi}$, reached at $x=\pi$.
A: Alternative proof
$f(x)=\cos(x)-1+\frac{2}{\pi}(x+\sin(x))$. Note that $f(0)=0, f(x) \to \infty$ when $x \to \infty$. We only need to show $f(x) \ge 0$ at every stationary point.
$$f'(x)=-\sin(x)+\frac{2}{\pi}(1+\cos(x))=-2\sin\frac x2 \cos\frac x2+\frac{2}{\pi}\cdot 2\cos^2 \frac x2\\=2\cos\frac x2\left(-\sin \frac x2 + \frac{2}{\pi}\cos \frac x2\right)$$
$$f(x)=-2\sin^2 \frac x2 + \frac{2}{\pi} x + \frac{2}{\pi} 2 \sin\frac x2 \cos \frac x2=\frac{2}{\pi}x + 2\sin\frac x2 \left(-\sin \frac x2+\frac{2}{\pi} \cos \frac x2\right)\\
$$
Now if $x_0$ is a stationary point, i.e., $f'(x_0)=0$, we have two cases:

*

*$\cos \frac{x_0}{2}=0$. Then $\sin \frac{x_0}{2} = 1, x_0 \ge \pi \implies f(x_0)=-2+\frac{2}{\pi}x_0 \ge 0.$

*$-\sin \frac{x_0}{2}+\frac{2}{\pi} \cos \frac{x_0}{2}=0$. Then $f(x_0) = \frac{2}{\pi}x_0 \ge 0.\blacksquare$
