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 2 of 5: Base case of the lemma

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A_1(x)+A_1(y)\ge A_0(xy)\iff xy(x-y)^2+(xy-1)^2\ge0

Detailed analysis

For $x,y>0$ , $A_1(x)+A_1(y)\ge A_0(xy)$ reads $\frac{2}{(1+x)^2}+\frac{2}{(1+y)^2}\ge\frac{2}{1+xy}$ . Clearing denominators, this is equivalent to $xy(x-y)^2+(xy-1)^2\ge0$ , which always holds, with equality iff $x=y=1$ .