# 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 $$t>0$$.

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 $$x=\pi$$.

• Indeed, the original function $1-\frac{2}{\pi}(x+\sin(x))$ is easily seen to be decreasing (we check its first derivative), which implies that the OP's inequality is clearly true for $x\ge\pi$ (and clearly false for $x<0$). This lovely argument can therefore concentrate on $[0,\pi]$. The OP should have specified the desired region of validity. Commented Dec 17, 2020 at 17:32

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 $$f(x_0) = \frac{2}{\pi}x_0 \ge 0.\blacksquare$$
• Really nice and straightforward. But I think there is a typo in the 1st term in 2nd step of evaluation of $f'(x)$ Commented Dec 18, 2020 at 5:54
• Thanks for the feedback. The first one was indeed a typo. The second one you still need $x_0 \ge \pi$. Commented Dec 18, 2020 at 13:36