BMO 2022 shortlist

Writing $x_{n+1}=x_1$, by AM-GM we have $$\frac{x_1+x_2}{x_2+x_3}+\frac{x_2+x_3}{x_3+x_4}+\cdots +\frac{x_{n-1}+x_n}{x_n+x_1}+\frac{x_n+x_1}{x_1+x_2}\ge n\sqrt{\prod _{i=1}^n\frac{x_i+x_{i+1}}{x_{i+1}+x_{i+1}}}=n.$$ So it is enough to prove that $$\frac{x_1}{x_1+x_2}+\frac{x_2}{x_2+x_3}+\cdots +\frac{x_n}{x_n+x_1}\ge \frac{n}{2}.$$ Letting $a_i=x_{i+1}/x_i$ for $i=1,2,\dots ,n$, it is enough to prove that $$\frac{1}{1+a_1}+\cdots +\frac{1}{1+a_n}\ge \frac{n}{2}.$$ Note that $a_1,\dots ,a_{n-1}\le 1$ and $a_1{a}_2\dots a_n=1$. 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 $$\frac{1}{1+a_1}+\cdots +\frac{1}{1+a_m}\ge \frac{m+1}{2}-\frac{a_1{a}_2\dots a_m}{1+a_1{a}_2\dots a_m}.$$ 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.

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Alternative solution.

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 $$\frac{1-a_1}{1+a_1}+\cdots +\frac{1-a_n}{1+a_n}\ge 0.$$ 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 $$\frac{1-a_1}{1+a_1}+\cdots +\frac{1-a_{n-1}}{1+a_{n-1}}\ge \frac{1-a_1\dots a_{n-1}}{1+a_1\dots a_{n-1}}=\frac{1-1/a_n}{1+1/a_n}=-\frac{1-a_n}{1+a_n}$$ and the result follows.