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 4 of 5: Chain the lemma along the product

Previous step Next step

A_i(a_i)+A_i(a_{i+1}\cdots a_n)\ge A_{i-1}(a_i\cdots a_n),\quad i=n,n-1,\ldots,1

Detailed analysis

Apply the lemma at level $i$ to the pair $x=a_i$ , $y=a_{i+1}\cdots a_n$ (with the empty product $a_{n+1}\cdots a_n$ read as $1$ ), for $i=n,n-1,\ldots,1$ . This produces $n$ inequalities $A_i(a_i)+A_i(a_{i+1}\cdots a_n)\ge A_{i-1}(a_i\cdots a_n)$ , whose right-hand side at level $i$ is exactly the second term on the left-hand side of the inequality at level $i-1$ .