Weighted AM-GM should work here. Then one can use ISL 1998 A.2 for $r_i=a_i+1$.

The solution CyclicISLscelesTrapezoid typed was the official solution whose motivation was telecoping the sum $$\dfrac{1}{2^n}+\sum_{i=1}^{n}{\dfrac{1}{2^i}\left(\dfrac{2}{1+a_i}\right)^{2^i}}$$ Note: Weighted AM-GM solution seems pretty natural at first sight, however after using that you should deal with $1-\dfrac{1}{2^n}$ in both coefficient and root as following: $$LHS\overbrace{\geq}^{Weighted} \left(1-\dfrac{1}{2^n}\right)\cdot \left(2\sum_{cyc}{\dfrac{1}{a_1+1}}\right)^{\dfrac{2^n}{2^n-1}}$$it still seems to be a good conjecture, however it is not a good choice because you should deal with $n$. Thus, telescoping the sum and proving like the official solution might be the best choice.

By the way, problem implies $$\sum_{i=1}^{n}{\dfrac{1}{2^i}\left(\dfrac{2}{i+a_i}\right)^{2^i}} \geq \dfrac{2}{\prod\limits_{i=1}^{n}{i}+a_1a_2\ldots a_n}-\dfrac{1}{2^n}$$ which is quite interesting. and... Who is the author ?

Funny problem, the hint is too obvious. We only need to prove for $\forall 1\le i\le n,$ $$\dfrac{1}{2^i}\left(\dfrac{2}{1+a_i}\right)^{2^i}\ge\dfrac{1}{2^{i-1}}\left(\dfrac{2}{1+a_ia_{i+1}\cdots a_n}\right)^{2^{i-1}}-\dfrac{1}{2^{i}}\left(\dfrac{2}{1+a_{i-1}a_{i}\cdots a_n}\right)^{2^{i}}$$and sum them up we are done. Let $x=a_i,y=a_{i+1}\cdots a_n,t=2^{i-1},$ $$\iff\left(\dfrac{2}{1+x}\right)^{2t}+\left(\dfrac{2}{1+y}\right)^{2t}\ge 2\left(\dfrac{2}{1+xy}\right)^{t}.$$It is not that easy if $t$ is a positive real number, but here simply induct on $i$ is enough. $$\left(\dfrac{2}{1+x}\right)^{2t}+\left(\dfrac{2}{1+y}\right)^{2t}\ge \frac 12\left(\left(\dfrac{2}{1+x}\right)^t+\left(\dfrac{2}{1+y}\right)^t\right)^2\ge 2\left(\dfrac{2}{1+xy}\right)^{t}$$by induction hypothesis of $i-1.\Box$

Many thanks to Seungjun_Lee for pointing out the error in my former solution.

Sammy27 wrote: But by Tchebycheff's inequality, we get that $$\frac{1+a_1a_2\dots a_{n-1}a_n}{2}\geq \frac{(1+a_1)(1+a_2)\dots (1+a_{n-1})(1+a_n)}{2^n},$$ What is Tchebycheff's inequality?

Sammy27 wrote: But by Tchebycheff's inequality, we get that $$\frac{1+a_1a_2\dots a_{n-1}a_n}{2}\geq \frac{(1+a_1)(1+a_2)\dots (1+a_{n-1})(1+a_n)}{2^n},$$ I don't think that this always holds. Try $n = 2$ then we get $(a_1-1)(a_2-1) \ge 0$

Nice inequality ! The idea is to guess the following lemma based on a well known result(the case $i=1$): $$(\frac{2}{1+a})^{2^i}+(\frac{2}{1+b})^{2^i}\ge 2\cdot (\frac{2}{1+ab})^{2^{i-1}}$$The induction step is trivial (use $a^2+b^2\ge \frac{(a+b)^2}{2}$) and the base case we want to prove that : $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\ge \frac{1}{1+ab}$ And for this use cauchy: $(1+a)^2\le (1+ab)(1+\frac{a}{b})$ and $(1+b)^2\le (1+ab)(1+\frac{b}{a})$ Now the problem is solved: use this inequalities add them and realise that you are done: For every $i=1,\dots n-1$ we have : $$\frac{1}{2^i}\cdot (\frac{2}{1+a_i})^{2^i}+\frac{1}{2^i}\cdot (\frac{2}{1+a_{i+1}\dots a_n})^{2^i}\ge \frac{1}{2^{i-1}}\cdot (\frac{2}{1+a_i\dots a_n})^{2^{i-1}}$$and this ineqaulity: $$\frac{1}{2^n}+\frac{1}{2^n}\cdot (\frac{2}{1+a_n})^{2^n}\ge \frac{1}{2^{n-1}}\cdot (\frac{2}{1+a_n})^{2^{n-1}} $$

Very amazing problem

a_507_bc wrote: Let $n$ be a positive integer and let $a_1, a_2, \ldots, a_n$ be positive reals. Show that $$\sum_{i=1}^{n} \frac{1}{2^i}(\frac{2}{1+a_i})^{2^i} \geq \frac{2}{1+a_1a_2\ldots a_n}-\frac{1}{2^n}.$$ For the contest collection, I think the LaTeX should be fixed like this: Let $n$ be a positive integer and $a_1,a_2,\ldots,a_n$ be positive real numbers. Prove that $$\sum_{i=1}^{n} \frac{1}{2^i} \left( \frac{2}{1+a_i} \right)^{2^i} \geq \frac{2}{1+a_1a_2\ldots a_n}-\frac{1}{2^n}.$$

My Solution.