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:
$$\frac{y_1}{y_1+1} + \frac{y_2}{y_2+1} + \dots + \frac{y_n}{y_n+1} \le n-1.$$
It is obvious that $y_1y_2\cdots y_n = 1$ and all $y_i$ are positive. Now we transform the last inequality:
$$n - \left( \frac{y_1}{y_1+1} + \frac{y_2}{y_2+1} + \dots + \frac{y_n}{y_n+1} \right)$$
$$1 - \frac{y_1}{y_1+1} + 1 - \frac{y_2}{y_2+1} + \dots + 1 - \frac{y_n}{y_n+1} \geq 1$$
$$\frac{1}{y_1+1} + \frac{1}{y_2+1} + \dots + \frac{1}{y_n+1} \geq 1$$
We will prove the last inequality by induction. For $n=3$ it is in the form $\frac{1}{y_1+1} + \frac{1}{y_2+1} + \frac{1}{y_3+1} \geq 1$ under the condition $y_1y_2y_3 = 1$. Now, $y_3 = \frac{1}{y_1y_2}$ and if we substitute this in the inequality we get: $\frac{1}{y_1+1} + \frac{1}{y_2+1} + \frac{y_1y_2}{1+y_1y_2} \geq 1$. Now if we use the inequality $\frac{1}{y_1+1} + \frac{1}{y_2+1} \geq \frac{1}{y_1y_2+1}$ (which is easily proved by multiplying the two sides) we get the required result.

Now let $\frac{1}{y_1+1} + \frac{1}{y_2+1} + \dots + \frac{1}{y_n+1} \geq 1$ hold for $n$. For $n+1$ we have
$$\frac{1}{y_1+1} + \frac{1}{y_2+1} + \dots + \frac{1}{y_{n+1}+1} \geq \frac{1}{y_1+1} + \frac{1}{y_2+1} + \dots + \frac{1}{y_{n-1}+1} + \frac{1}{1+y_n} + \frac{1}{1+y_{n+1}} \geq \\ \geq \frac{1}{y_1+1} + \frac{1}{y_2+1} + \dots + \frac{1}{y_{n-1}+1} + \frac{1}{1+y_n y_{n+1}}$$
where we used $\frac{1}{y_n+1} + \frac{1}{y_{n+1}+1} \geq \frac{1}{y_n y_{n+1}+1}$ in a similar way as above. So now because $y_1y_2 \cdots y_{n+1} = 1$ if we denote $t_i = y_i$ for $i = 1, \dots, n-1$ and $t_n = y_n y_{n+1}$ we have $t_1t_2 \cdots t_n = 1$ and the last inequality is $\frac{1}{1+t_1} + \frac{1}{1+t_2} + \dots + \frac{1}{1+t_n}$ which is true according to the inductive assumption.