Mongolian Mathematical Olympiad

By Taylor's theorem, we have $$\sqrt{1-t}=1-\sum _{k=1}^{\infty }\frac{(2k)!}{4^k(k!{)}^2(2k-1)}{t}^k$$ for any $-1<t<1$. Since ${\sum }_{i=1}^n{x}_i=0$, we have $$\sum _{i=1}^n\sum _{j=1}^n{x}_i{x}_j\sqrt{1-x_i^2{x}_j^2}={\left(\sum _{i=1}^n{x}_i\right)}^2-\sum _{k=1}^{\infty }\frac{(2k)!}{4^k(k!{)}^2(2k-1)}{\left(\sum _{i=1}^n{x}_i^{2k+1}\right)}^2\le 0.$$ Equality holds if and only if ${\sum }_{i=1}^n{x}_i^{2k+1}=0$ for all $k\ge 1$. This means that the list $x_1,x_2,\dots ,x_n$ consists of zeros and opposite numbers. Indeed, equality holds for such numbers. Now suppose ${\sum }_{i=1}^n{x}_i^{2k+1}=0$ for all $k\ge 1$. Removing the zeros, and negating the negative numbers, we get positive numbers $a_1,a_2,\dots ,a_m>0$ and $b_1,b_2,\dots ,b_l>0$ such that ${\sum }_i{a}_i^{2k+1}={\sum }_j{b}_j^{2k+1}$ for all $k\ge 1$. It is easy to see that this implies $\max \{a_i\}=\max \{b_j\}$ considering a large enough $k$. By induction, the list consists of zeros and opposite numbers.