IMO 1996 SL A3

Origin: GRE | Category: Algebra

Problem

Let $a>2$ be given, and define recursively

$$a_0 = 1, \quad a_1 = a, \quad a_{n+1} = \left(\frac{a_n^2}{a_{n-1}^2}-2\right)a_n.$$

Show that for all $k \in \mathbb{N}$, we have

$$\frac{1}{a_0+1} + \frac{1}{a_1+1} + \frac{1}{a_2+1} + \cdots + \frac{1}{a_k+1} < \frac{1}{2} + a - \sqrt{a^2-4}.$$

Solution

Since $a_1>2$, it can be written as $a_1 = b + b^{-1}$ for some $b>0$. Furthermore,

$$a_1^2 - 2 = b^2 + b^{-2} \quad \text{and hence} \quad a_2 = (b^2 + b^{-2})(b + b^{-1}).$$

We prove that

$$a_n = (b+b^{-1})(b^2+b^{-2})(b^4+b^{-4}) \cdots (b^{2^{n-1}} + b^{-2^{n-1}})$$

by induction. Indeed,

$$\frac{a_{n+1}}{a_n} = \frac{a_n}{a_{n-1}} - 2 = \frac{b^{2^n}+b^{-2^n}}{1}.$$

Now we have

$$\sum_{i=0}^n \frac{1}{a_i+1} = 1 + \frac{b}{b^2+1} + \frac{b^3}{(b^2+1)(b^4+1)} + \cdots + \frac{b^{2^{n}-1}}{(b^2+1)(b^4+1)\cdots (b^{2^n}+1)}. \tag{1}$$

Note that

$$\frac{1}{2}\left(a+2-\sqrt{a^2-4}\right) = 1 + \frac{1}{b},$$

hence we must prove that the right side in (1) is less than $1/b$. This follows from the fact that

$$\frac{b^{2^k}}{(b^2+1)(b^4+1)\cdots (b^{2^k}+1)} = \frac{(b^2+1)(b^4+1)\cdots (b^{2^{k-1}}+1) - (b^2+1)(b^4+1)\cdots (b^{2^k}+1)}{1},$$

hence the right side in (1) equals

$$\frac{1}{b}\left(1 - \frac{(b^2+1)(b^4+1)\cdots (b^{2^n}+1)}{1}\right),$$

and this is clearly less than $1/b$.