North Macedonia

In every one of the fractions $\frac{x_{i-1}{x}_{i+1}}{x_{i-1}{x}_{i+1}+x_i{x}_{i+2}}$ (where $i\in \{1,2,\dots ,n\}$ and $x_0=x_n$ and $x_{n+1}=x_1$) we divide by $x_i{x}_{i+2}$ and if we denote $y_i=\frac{x_{i-1}{x}_{i+1}}{x_i{x}_{i+2}}$ the required inequality is transformed to the form: $$\frac{y_1}{y_1+1}+\frac{y_2}{y_2+1}+\cdots +\frac{y_n}{y_n+1}\le n-1.$$ It is obvious that $y_1{y}_2\cdots y_n=1$ and all $y_i$ are positive. Now we transform the last inequality: $$n-\left(\frac{y_1}{y_1+1}+\frac{y_2}{y_2+1}+\cdots +\frac{y_n}{y_n+1}\right)$$ $$1-\frac{y_1}{y_1+1}+1-\frac{y_2}{y_2+1}+\cdots +1-\frac{y_n}{y_n+1}\ge 1$$ $$\frac{1}{y_1+1}+\frac{1}{y_2+1}+\cdots +\frac{1}{y_n+1}\ge 1$$ We will prove the last inequality by induction. For $n=3$ it is in the form $\frac{1}{y_1+1}+\frac{1}{y_2+1}+\frac{1}{y_3+1}\ge 1$ under the condition $y_1{y}_2{y}_3=1$. Now, $y_3=\frac{1}{y_1{y}_2}$ and if we substitute this in the inequality we get: $\frac{1}{y_1+1}+\frac{1}{y_2+1}+\frac{y_1{y}_2}{1+y_1{y}_2}\ge 1$. Now if we use the inequality $\frac{1}{y_1+1}+\frac{1}{y_2+1}\ge \frac{1}{y_1{y}_2+1}$ (which is easily proved by multiplying the two sides) we get the required result.

Now let $\frac{1}{y_1+1}+\frac{1}{y_2+1}+\cdots +\frac{1}{y_n+1}\ge 1$ hold for $n$. For $n+1$ we have $$\begin{gathered} \frac{1}{y_1+1}+\frac{1}{y_2+1}+\cdots +\frac{1}{y_{n+1}+1}\ge \frac{1}{y_1+1}+\frac{1}{y_2+1}+\cdots +\frac{1}{y_{n-1}+1}+\frac{1}{1+y_n}+\frac{1}{1+y_{n+1}}\ge \\ \ge \frac{1}{y_1+1}+\frac{1}{y_2+1}+\cdots +\frac{1}{y_{n-1}+1}+\frac{1}{1+y_n{y}_{n+1}} \end{gathered}$$ where we used $\frac{1}{y_n+1}+\frac{1}{y_{n+1}+1}\ge \frac{1}{y_n{y}_{n+1}+1}$ in a similar way as above. So now because $y_1{y}_2\cdots y_{n+1}=1$ if we denote $t_i=y_i$ for $i=1,\dots ,n-1$ and $t_n=y_n{y}_{n+1}$ we have $t_1{t}_2\cdots t_n=1$ and the last inequality is $\frac{1}{1+t_1}+\frac{1}{1+t_2}+\cdots +\frac{1}{1+t_n}$ which is true according to the inductive assumption.