Problem 2413

Writing $x_{n+1} = x_1$, by AM-GM we have
$$\frac{x_1+x_2}{x_2+x_3} + \frac{x_2+x_3}{x_3+x_4} + \dots + \frac{x_{n-1}+x_n}{x_n+x_1} + \frac{x_n+x_1}{x_1+x_2} \ge n \sqrt{\prod_{i=1}^{n} \frac{x_i+x_{i+1}}{x_{i+1}+x_{i+1}}} = n.$$
So it is enough to prove that
$$\frac{x_1}{x_1+x_2} + \frac{x_2}{x_2+x_3} + \dots + \frac{x_n}{x_n+x_1} \ge \frac{n}{2}.$$
Letting $a_i = x_{i+1}/x_i$ for $i = 1, 2, \dots, n$, it is enough to prove that
$$\frac{1}{1+a_1} + \dots + \frac{1}{1+a_n} \ge \frac{n}{2}.$$
Note that $a_1, \dots, a_{n-1} \le 1$ and $a_1a_2 \dots a_n = 1$.
Equivalently, it is enough to prove that if $m \ge 2$ is an integer and $a_1, \dots, a_m \le 1$ are real numbers then
$$\frac{1}{1+a_1} + \dots + \frac{1}{1+a_m} \ge \frac{m+1}{2} - \frac{a_1 a_2 \dots a_m}{1+a_1 a_2 \dots a_m}.$$
We proceed by induction on $m$. In fact the statement is true even for $m = 1$ so we assume that it is true for $m = k$ and proceed with the inductive step. Letting $a = a_1 \dots a_k$ and $b = a_{k+1}$ it is enough to prove that
$$\frac{1}{1+b} - \frac{a}{1+a} \ge \frac{1}{2} - \frac{ab}{1+ab}.$$
We have
$$\frac{a}{1+a} - \frac{ab}{1+ab} = \frac{a(1-b)}{(1+a)(1+ab)} \le \frac{1-b}{1+b} = \frac{1}{1+b} - \frac{1}{2}$$
so the result follows.

Since $\frac{1}{1+x} = \frac{1}{2} + \frac{1}{2} \cdot \frac{1-x}{1+x}$, with the notation of Solution 1 it is enough to prove that
$$\frac{1-a_1}{1+a_1} + \dots + \frac{1-a_n}{1+a_n} \ge 0.$$
Letting $f(x) = \frac{1-x}{1+x}$ one can check that
$$f(x) + f(y) - f(xy) = \frac{(1-x)(1-y)(1-xy)}{(1+x)(1+y)(1+xy)}.$$
Thus $f(x) + f(y) \ge f(xy)$ for $x, y \le 1$. So inductively
$$\frac{1-a_1}{1+a_1} + \dots + \frac{1-a_{n-1}}{1+a_{n-1}} \ge \frac{1-a_1 \dots a_{n-1}}{1+a_1 \dots a_{n-1}} = \frac{1-1/a_n}{1+1/a_n} = -\frac{1-a_n}{1+a_n}$$
and the result follows.