Prove $\sum_{k=1}^{n}\sqrt{x^2+2x\cos\frac{(2k-1)\pi}{2n} + 1}\ge 1+ \sum_{k=1}^{n-1} \sqrt{x^2+2x\cos\frac{2k\pi}{2n} + 1}$

Question

For nonnegative real number $x \ge 0$, and positive integer $n>0$,prove that $$ \sum_{k=1}^n\sqrt{x^2+2x\cos\frac{(2k-1)\pi}{2n} + 1} \ge 1+ \sum_{k=1}^{n-1} \sqrt{x^2+2x\cos\frac{2k\pi}{2n} + 1}$$ where equality holds when $x=0$.

(PS: I am not sure whether the inequality holds for $0\le x \le cos\frac{\pi}{2n}$)

(The idea is from here: https://arxiv.org/abs/cond-mat/0509490, left down corner of page 2, "The parity of the number of c-quasiparticles is a good quantum number and the ground state has even parity for any value of g.")

【newly updated:】 Let $y = f(x)= \sum_{k=1}^n\sqrt{x^2+2x\cos\frac{(2k-1)\pi}{2n} + 1} -( 1+ \sum_{k=1}^{n-1} \sqrt{x^2+2x\cos\frac{2k\pi}{2n} + 1})$,

for 2n=100 and 2n=200, draw f(x)'s graphics with mathematica, as follows: the inequality holds for $ x \ge cos\frac{\pi}{2n}$,

So, the question is: in interval[0,$cos\frac{\pi}{2n}$), inequality holds with computer's numerical error or not holds?

Answer 1

Hints:

1. When $z=e^{i\phi}$ we have $|x+z|=\sqrt{x^2+2x\cos\phi+1}$.
2. When $x=0$ the claimed inequality holds with equality.
3. If $0<\alpha<\beta<\pi/2$ then $|x+e^{i\alpha}|\ge |x+e^{i\beta}|$ for all $x\ge0$. Draw a picture if you don't see why this is true!