This is a not hard but elegant problem. Define $s=\sum a_i\ge 0$. If $s\ge1$,it's obvious. If $0\le s \le 1$, then $$(1-\sum a_i\cos x_i)^2+(1-\sum a_i\sin x_i)^2\ge \dfrac12 (2-\sum a_i(\cos x_i+\sin x_i))^2\ge (\sqrt{2}-s)^2$$Thus we only need to prove $(\sqrt2-s)^4\ge 4(1-s)^3\Longleftarrow 4(3-2\sqrt2)+s^3\ge 4(\sqrt2-1)s^2$. By AM-GM inequality,$$4(\sqrt{2}-1)s^2-s^3=4\cdot\frac{s}2\cdot\frac{s}2\cdot(4(\sqrt{2}-1)-s)\le 4(\dfrac{4(\sqrt{2}-1)}3)^3$$We only need to prove $4(\dfrac{4(\sqrt{2}-1)}3)^3\le 4(3-2\sqrt2)\Longleftrightarrow \sqrt{2}-1\le \frac{27}{64}=0.421875$.which is true.

$(\sqrt2-s)^4\ge 4(1-s)^3\Longleftarrow 4(3-2\sqrt2)+s^3\ge 4(\sqrt2-1)s^2$? $(\sqrt2-s)^2\ge 4(1-s)^3\Longleftarrow 4(3-2\sqrt2)+s^3\ge 4(\sqrt2-1)s^2$

envision2017 wrote: $(\sqrt2-s)^4\ge 4(1-s)^3\Longleftarrow 4(3-2\sqrt2)+s^3\ge 4(\sqrt2-1)s^2$? $(\sqrt2-s)^2\ge 4(1-s)^3\Longleftarrow 4(3-2\sqrt2)+s^3\ge 4(\sqrt2-1)s^2$ I'm correct. $(\sqrt2-s)^4\ge 4(1-s)^3\iff 4(3-2\sqrt2)s+s^4\ge 4(\sqrt2-1)s^3\Longleftarrow 4(3-2\sqrt2)+s^3\ge 4(\sqrt2-1)s^2$

smy2012 wrote: This is a not hard but elegant problem. Define $s=\sum a_i\ge 0$. If $s\ge1$,it's obvious. If $0\le s \le 1$, then $$(1-\sum a_i\cos x_i)^2+(1-\sum a_i\sin x_i)^2\ge \dfrac12 (2-\sum a_i(\cos x_i+\sin x_i))^2\ge (\sqrt{2}-s)^2$$Thus we only need to prove $(\sqrt2-s)^4\ge 4(1-s)^3\Longleftarrow 4(3-2\sqrt2)+s^3\ge 4(\sqrt2-1)s^2$. By AM-GM inequality,$$4(\sqrt{2}-1)s^2-s^3=4\cdot\frac{s}2\cdot\frac{s}2\cdot(4(\sqrt{2}-1)-s)\le 4(\dfrac{4(\sqrt{2}-1)}3)^3$$We only need to prove $4(\dfrac{4(\sqrt{2}-1)}3)^3\le 4(3-2\sqrt2)\Longleftrightarrow \sqrt{2}-1\le \frac{27}{64}=0.421875$.which is true. How did the latter part of the 4th line came to be?

The equality holds when $x_i=\frac{\pi}{4}+2k\pi$ and $a_i=0$ where $i=1,2,\ldots ,n$ and $k$ is an integer. By Cauchy-Schwarz Inequality, we have \[(1^2+1^2)^2\left(\left(1-\sum_{i=1}^n a_i\cos x_i\right)^2+\left(1-\sum_{i=1}^n a_i\sin x_i\right)^2\right)^2\ge \left(2-\sum_{i=1}^n a_i (\cos x_i +\sin x_i)\right)^4=\left(2-\sqrt 2\sum_{i=1}^n a_i \sin \left(\frac{\pi}{4}+x_i\right)\right)^4 \ge 4 \left(\sqrt 2-\sum_{i=1}^n a_i \right)^4.\]So, it is equivalent to show that $\left(\sqrt 2-\sum_{i=1}^n a_i \right)^4\geq 4(1-\sum_{i=1}^n a_i)^3$. Let $t=\sum_{i=1}^n a_i $, it is equivalent to showing \[t(t^3-4\sqrt 2t^2+4t^2-8\sqrt 2+12) \geq 0\]since $t\geq 0$, let $f(t)=t^3-4\sqrt 2t^2+4t^2-8\sqrt 2+12$, we have to check that $f(t)\geq 0$ for all $t\geq 0$, so $f'(t)=t(3t-8\sqrt2+8)=0 \implies t=0, \frac{8\sqrt 2 -8}{3}$ are the min/max points. It remains to check that $f\left(\frac{8\sqrt 2 -8}{3}\right) \geq 0$. $\blacksquare$