# Inequality, triangle, Law of cosines， integer

Prove $$|a^2+1-2a\cos{\theta}|^{\frac{1}{n}}\ge| a^{\frac{2}{n}}+1-2a^{\frac{1}{n}}\cos{\frac{\theta}{n}}|$$ where $a>0$ , $0<\theta<\pi$ and $n\ge2$ and $n\in N^+$.

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Write $$\cos x = \frac{1}{2} (e^{ix} + e^{-ix})$$ Both sides look like a variation of the binomial formula. –  AlexR Sep 13 '13 at 11:39
The inequality is equivalent to $$|z^n-1|\ge|z-1|^n$$, where $z$ is a complex number and $\arg{z}\in(0,\frac{\pi}{n})$.
Define the solution of $z^n-1=0$ by $\varsigma^k=\cos{\frac{2k\pi}{n}}+i\sin{\frac{2k\pi}{n}}$, where $k=0,1,2\cdots,n-1$.
We have $$z^n-1=(z-\varsigma)(z-\varsigma^2)\cdots(z-\varsigma^{n-1}).$$ Now, what we need to do is to prove $|(z-\varsigma^k)|\ge |z-1|$. In the view of the geometric meaning of it, we see that $$|(z-\varsigma^k)|^2=|z|^2+1-2\cos{\big<z,\varsigma^k\big>}\ge |z|^2+1-2\cos{\frac{\pi}{n}}$$ and $$|(z-1)|^2=|z|^2+1-2\cos{\big<z,1\big>}\le|z|^2+1-2\cos{\frac{\pi}{n}}.$$ It completes our proof.