Proving the lower bound $|1-e^{i\theta}| \ge 2|\theta|/\pi$

Question

I came upon a text using the lower bound $$|1-e^{i\theta}| \ge 2|\theta|/\pi$$ which should hold whenever $$-\pi\le\theta\le\pi$$ I have no idea how to prove it. I thought about using the Taylor expansion, but don't see how to proceed from here; I'd like to know also if there's a standard technique I'm missing

Answer 1

It is $|1-e^{i\theta}|=2|\sin \theta/2|$. Assume $0 \le \theta \le \pi$, hence $0\le\theta/2 \le \pi/2$; the function $\sin x$ is concave for $x\in[0,\pi/2]$, and so it is $\sin x \ge \frac{2}{\pi}x$ for each $x \in [0,\pi/2]$. This is enough to conclude using the concavity inequality with $x=\theta/2$.

Since your inequality is symmetric by exchanging $\theta$ with $-\theta$, you get the inequality with the modulus on the right hand side.

Answer 2

I like your question, and here is a geometric intuition (sorry for not drawing anything, I just imagined the proof :-) )

The largest Euclidean distance between two points on a circle with radius of $1$ is $2$ and these points have the larges distance along the circle (which is $\pi$). Now note that if you choose two points on a circle and move one away from the other, then the difference between their Euclidean distance and the corresponding sector grows strictly from $0$ (when points fall on each other) to $\pi-2$ [so that the inequality you are after will follow].