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 1 of 5: Restate the inequality

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A_k(x)=\frac{1}{2^k}\left(\frac{2}{1+x}\right)^{2^k},\qquad\sum_{i=1}^nA_i(a_i)\ge A_0(a_1a_2\cdots a_n)-\frac{1}{2^n}

Detailed analysis

Write $A_k(x)=\frac{1}{2^k}\left(\frac{2}{1+x}\right)^{2^k}$ for $k\ge0$ , so that $A_0(x)=\frac{2}{1+x}$ . The desired inequality becomes $\sum_{i=1}^nA_i(a_i)\ge A_0(a_1a_2\cdots a_n)-\frac{1}{2^n}$ .