IMO 1982 · Problem 3

Consider the infinite sequences \(\{x_n\}\) of positive real numbers with the following properties:

\[ x_0=1,\text{ and for all }i\ge0,\ x_{i+1}\le x_i. \]

(a) Prove that for every such sequence, there is a \(n\ge1\) such that

\[ \frac{x_0^2}{x_1}+\frac{x_1^2}{x_2}+\cdots+\frac{x_{n-1}^2}{x_n}\ge3.999. \]

(b) Find such a sequence for which

\[ \frac{x_0^2}{x_1}+\frac{x_1^2}{x_2}+\cdots+\frac{x_{n-1}^2}{x_n}<4. \]

Official text of the problem. Official problem paper (PDF)

Answer

For (b), take \(\boxed{x_n=2^{-n}\quad(n\ge0)}\). Its indicated partial sum is \(4(1-2^{-n})<4\) for every \(n\ge1\).

Solution

(a) Write \[ S_n=\sum_{i=0}^{n-1}\frac{x_i^2}{x_{i+1}}. \] For positive \(a,b\), \[ \frac{a^2}{b}-4(a-b)=\frac{(a-2b)^2}{b}\ge0. \] Applying this with \(a=x_i\), \(b=x_{i+1}\), and summing gives \[ S_n\ge4\sum_{i=0}^{n-1}(x_i-x_{i+1}) =4(1-x_n). \]

If \(x_n\le1/4000\) for some \(n\ge1\), then \[ S_n\ge4\left(1-\frac1{4000}\right)=3.999. \]

Otherwise \(x_i>1/4000\) for every \(i\ge0\). Since \(x_{i+1}\le x_i\), \[ \frac{x_i^2}{x_{i+1}}\ge x_i>\frac1{4000}. \] Consequently, \[ S_{15996}>\frac{15996}{4000}=3.999. \] Thus the required \(n\) exists in either case.

(b) Set \(x_n=2^{-n}\). This sequence is positive, satisfies \(x_0=1\), and is decreasing. For every \(n\ge1\), \[ S_n =\sum_{i=0}^{n-1}\frac{2^{-2i}}{2^{-(i+1)}} =\sum_{i=0}^{n-1}2^{1-i} =4(1-2^{-n}) <4. \] Hence it satisfies all the requirements.

The solution was checked against John Scholes' IMO solution archive (Kalva).