Problem 3

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}\ge\frac{2}{1+a_1a_2\cdots a_n}-\frac{1}{2^n}.

Step 5 of 5: Telescope and conclude

Previous step Next step

\sum_{i=1}^nA_i(a_i)+\frac{1}{2^n}\ge A_0(a_1a_2\cdots a_n)

Detailed analysis

Summing the $n$ inequalities of the previous step, every shared term cancels between consecutive levels except the first term of each left-hand side and the term $A_n(1)=1/2^n$ from the top level; this leaves $\sum_{i=1}^nA_i(a_i)+\frac{1}{2^n}\ge A_0(a_1a_2\cdots a_n)$ , which is exactly $\sum_{i=1}^n\frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i}\ge\frac{2}{1+a_1a_2\cdots a_n}-\frac{1}{2^n}$ , with equality iff $a_1=a_2=\cdots=a_n=1$ .