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Serbian Mathematical Olympiad

Algebra Difficulty 4.6 AIME Prove it Serbia

Let $n \geqslant 2$ be a natural number and let positive real numbers $a_{0}, a_{1}, \ldots, a_{n}$ satisfy the equality
$\left(a_{k-1}+a_{k}\right)\left(a_{k}+a_{k+1}\right)=a_{k-1}-a_{k+1} \quad \text{ for every } k=1,2, \ldots, n-1 \text{. }$
Prove that $a_{n}<\frac{1}{n-1}$ .

Solution

Solution:

The given equality is equivalent to
$\frac{1}{a_{k}+a_{k+1}}=1+\frac{1}{a_{k-1}+a_{k}}$
for every $k>0$ . By induction it follows that $\frac{1}{a_{k}+a_{k+1}}=k+\frac{1}{a_{0}+a_{1}}$ for $k>0$ , from which we obtain that $\frac{1}{a_{n-1}+a_{n}} > n-1$ and therefore $a_{n}<\frac{1}{n-1}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from sr; metadata (topic, difficulty) added by this project.