A problem on inequality and number theory Suppose $n \in \mathbb{Z}^+$; 
$$\sum_{i=1}^n\sqrt{1-x_ix_{i+1}}\ge\sqrt{n(n-1)}\tag{1}$$
where $,  x_1= x_{n+1}, \displaystyle \quad\forall x_i\in\mathbb{R} \text{ and }\sum_{i=1}^nx_i^2=1 $
Is it possible to find all the positive integer numbers $n$ which satisfy $(1)$ ?
 A: A proof for $n=4$.
Id est, for $a^2+b^2+c^2+d^2=1$ we need to prove that:
$$\sqrt{1-ab}+\sqrt{1-bc}+\sqrt{1-cd}+\sqrt{1-da}\geq2\sqrt3$$ or
$$\sum\limits_{cyc}(1-ab)+2\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+2\sqrt{(1-ab)(1-cd)}+2\sqrt{(1-bc)(1-da)}\geq12$$ or
$$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+\sqrt{(1-ab)(1-cd)}+\sqrt{(1-bc)(1-da)}\geq4+\frac{1}{2}\sum\limits_{cyc}ab$$
Now by C-S we obtain
$$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}=\sum\limits_{cyc}\sqrt{(a^2-ab+b^2+c^2+d^2)(b^2-bc+c^2+a^2+d^2)}=$$
$$=\sum\limits_{cyc}\sqrt{\left(\left(b-\frac{a}{2}\right)^2+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}c^2+d^2\right)\left(\left(b-\frac{c}{2}\right)^2+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}a^2+d^2\right)}\geq$$
$$\sum\limits_{cyc}\left(\left(b-\frac{a}{2}\right)\left(b-\frac{c}{2}\right)+\frac{3}{4}a^2+\frac{3}{4}c^2+\frac{1}{4}ac+d^2\right)=$$
$$=\frac{7}{2}-ab-bc-cd-da+ac+bd$$
In another hand, by C-S again we obtain
$$\sqrt{(1-ab)(1-cd)}=\sqrt{(a^2-ab+b^2+c^2+d^2)(c^2-cd+d^2+a^2+b^2)}=$$
$$=\sqrt{\left(\frac{(a-b)^2}{2}+\frac{1}{2}+\frac{1}{2}c^2+\frac{1}{2}d^2\right)\left(\frac{(c-d)^2}{2}+\frac{1}{2}+\frac{1}{2}a^2+\frac{1}{2}b^2\right)}\geq$$
$$\geq\frac{(a-b)(d-c)}{2}+\frac{1}{2}+\frac{1}{2}(ad+bc)=\frac{1}{2}+ad+bc-\frac{1}{2}(ac+bd)$$
similarly we'll obtain
$$\sqrt{(1-ad)(1-bc)}\geq\frac{1}{2}+ab+cd-\frac{1}{2}(ac+bd)$$
Thus, 
$$\sum\limits_{cyc}\sqrt{(1-ab)(1-bc)}+\sqrt{(1-ab)(1-cd)}+\sqrt{(1-bc)(1-da)}\geq\frac{9}{2}$$
and it remains to prove that
$$\frac{9}{2}\geq4+\frac{1}{2}\sum\limits_{cyc}ab$$ or
$$a^2+b^2+c^2+d^2\geq ab+bc+cd+da$$ or
$$(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2\geq0$$
Done!
