The Problems of Ukrainian Authors

Let's denote the left side of inequality that should be proven as $S$. Let $x_{14}{x}_1$ be the smallest product among all pairwise products of adjacent numbers $x_k{x}_{k+1}$, $k=1,14$ (cyclically adjacent numbers: $x_{15}=x_1$). Then we can bound from above all the summands containing this pair like this: $$x_{12}{x}_{13}{x}_{14}{x}_1\le x_{12}{x}_{13}{x}_6{x}_7,x_{13}{x}_{14}{x}_1{x}_2\le x_{13}{x}_7{x}_8{x}_2,x_{14}{x}_1{x}_2{x}_3\le x_8{x}_9{x}_2{x}_3,$$ But then $$\begin{array}{rl}S& =x_1{x}_2{x}_3{x}_4+x_2{x}_3{x}_4{x}_5+\cdots +x_{11}{x}_{12}{x}_{13}{x}_{14}+x_{12}{x}_{13}{x}_{14}{x}_1+x_{13}{x}_{14}{x}_1{x}_2+x_{14}{x}_1{x}_2{x}_3\\ & \le x_1{x}_2{x}_3{x}_4+x_2{x}_3{x}_4{x}_5+\cdots +x_{11}{x}_{12}{x}_{13}{x}_{14}+x_{12}{x}_{13}{x}_6{x}_7+x_{13}{x}_7{x}_8{x}_2+x_8{x}_9{x}_2{x}_3=P.\end{array}$$

In every group of $4$ summands from $P$ every multiplier has different remainder modulo $4$, that's why $$\begin{array}{rl}& (x_1+x_5+x_9+x_{13})(x_2+x_6+x_{10}+x_{14})(x_3+x_7+x_{11})(x_4+x_8+x_{12})\le \\ & \le \frac{1}{4^4}{\left(\sum _{k=1}^{14}{x}_k\right)}^4=\frac{1}{4^4},\end{array}$$ what was to be demonstrated.