APMO

We first prove the following lemma: Lemma 1. For $k$ positive integer and $x,y>0$, $${\left(\frac{2}{1+x}\right)}^{2^k}+{\left(\frac{2}{1+y}\right)}^{2^k}\ge 2{\left(\frac{2}{1+xy}\right)}^{2^{k-1}}.$$ The proof goes by induction. For $k=1$, we have $${\left(\frac{2}{1+x}\right)}^2+{\left(\frac{2}{1+y}\right)}^2\ge 2\left(\frac{2}{1+xy}\right)$$ which reduces to $$xy(x-y{)}^2+(xy-1{)}^2\ge 0.$$ For $k>1$, by the inequality $2\left(A^2+B^2\right)\ge (A+B{)}^2$ applied at $A={\left(\frac{2}{1+x}\right)}^{2^{k-1}}$ and $B={\left(\frac{2}{1+y}\right)}^{2^{k-1}}$ followed by the induction hypothesis $$\begin{array}{rl}2\left({\left(\frac{2}{1+x}\right)}^{2^k}+{\left(\frac{2}{1+y}\right)}^{2^k}\right)& \ge {\left({\left(\frac{2}{1+x}\right)}^{2^{k-1}}+{\left(\frac{2}{1+y}\right)}^{2^{k-1}}\right)}^2\\ & \ge {\left(2{\left(\frac{2}{1+xy}\right)}^{2^{k-2}}\right)}^2=4{\left(\frac{2}{1+xy}\right)}^{2^{k-1}}\end{array}$$ from which the lemma follows. The problem now can be deduced from summing the following applications of the lemma, multiplied by the appropriate factor: $$\begin{array}{rl}\frac{1}{2^n}{\left(\frac{2}{1+a_n}\right)}^{2^n}+\frac{1}{2^n}{\left(\frac{2}{1+1}\right)}^{2^n}& \ge \frac{1}{2^{n-1}}{\left(\frac{2}{1+a_n\cdot 1}\right)}^{2^{n-1}}\\ \frac{1}{2^{n-1}}{\left(\frac{2}{1+a_{n-1}}\right)}^{2^{n-1}}+\frac{1}{2^{n-1}}{\left(\frac{2}{1+a_n}\right)}^{2^{n-1}}& \ge \frac{1}{2^{n-2}}{\left(\frac{2}{1+a_{n-1}{a}_n}\right)}^{2^{n-2}}\\ \frac{1}{2^{n-2}}{\left(\frac{2}{1+a_{n-2}}\right)}^{2^{n-2}}+\frac{1}{2^{n-2}}{\left(\frac{2}{1+a_{n-1}{a}_n}\right)}^{2^{n-2}}& \ge \frac{1}{2^{n-3}}{\left(\frac{2}{1+a_{n-2}{a}_{n-1}{a}_n}\right)}^{2^{n-3}}\\ \dots & 2^{2^k}\\ \frac{1}{2^k}{\left(\frac{2}{1+a_k}\right)}^{2^k}+\frac{1}{2^k}{\left(\frac{2}{1+a_{k+1}\dots a_{n-1}{a}_n}\right)}^{2^{k-1}}& \ge \frac{1}{2^{k-1}}{\left(\frac{2}{1+a_k\dots a_{n-2}{a}_{n-1}{a}_n}\right)}^n\\ \frac{1}{2}{\left(\frac{2}{1+a_1}\right)}^2+\frac{1}{2}{\left(\frac{2}{1+a_2\dots a_{n-1}{a}_n}\right)}^2& \ge \frac{2}{1+a_1\dots a_{n-2}{a}_{n-1}{a}_n}.\end{array}$$