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1982 IMO Problems/Problem 3

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Problem

Consider infinite sequences $\{x_n\}$ of positive reals such that $x_0=1$ and $x_0\ge x_1\ge x_2\ge\ldots$.

a) Prove that for every such sequence there is an $n\ge1$ such that:\[{x_0^2\over x_1}+{x_1^2\over x_2}+\ldots+{x_{n-1}^2\over x_n}\ge3.999.\]

b) Find such a sequence such that for all $n$:\[{x_0^2\over x_1}+{x_1^2\over x_2}+\ldots+{x_{n-1}^2\over x_n}<4.\]

Solution 1

By Cauchy, the LHS is at least: $\frac{(x_{0}+...+x_{n-1})^{2}}{x_{1}+...+x_{n}}.$

It is clear that $x_{0}=1$ and $x_{n}\le \frac{x_{1}+...+x_{n-1}}{n-1}$

We thus have that the LHS is at least: $(\frac{1+2(x_{1}+...+x_{n-1})+(x_{1}+...+x_{n-1})^{2}}{x_{1}+...+x_{n-1}})(\frac{n-1}{n})$

Applying 2-variable AM-GM to $1$ and $(x_{1}+...+x_{n-1})^{2}$ the LHS is at least $4(\frac{n-1}{n})$

We then simply choose $n$ such that $4(\frac{n-1}{n}) \ge 3.999$.

Part b) is likely to occur only if both the AM-GM and Cauchy equality conditions are close to met, or if $1=x_{1}+x_{2}+...$ and $\frac{x_{0}}{x_{1}}=\frac{x_{1}}{x_{2}}=...$, respectively. This points to the geometric series with common ratio $\frac{1}{2}$ as the equality case.

This solution was posted and copyrighted by n^4 4. The original thread for this problem can be found here: [1]

Solution 2

b) is immediate by taking $x_i = 2^{-i}$.

For a), we prove $LHS \ge 2^{2^0 + 2^{-1}+...+ 2^{2-n}}$, and taking $n\to \infty$ will yield the result. (Here $n=1$ gives the empty sum, so $LHS \ge 2^0=1$, which is trivial).

The proof is not hard. Let $s_i$ be the sum of the first $i$ terms, so $s_n = s_{n-2} + \dfrac{x_{n-2}^2}{x_{n-1}} + \dfrac{x_{n-1}^2}{x_n} \ge s_{n-2} + 2^{2^0} x_{n-2} \sqrt{\dfrac{x_{n-1}}{x_n}} \ge s_{n-2} + 2^{2^0} x_{n-2}$. This in turn is $s_{n-3} + \dfrac{x_{n-3}^2}{x_{n-2}} + 2^{2^0} x_{n-2}\ge s_{n-3} + 2^{2^0 + 2^{-1}} x_{n-3}$. By repeatedly applying AM-GM to the last two terms of the sum, collecting powers of two, and cancelling variables, we eventually obtain

$s_0 + \dfrac{x_0^2}{x_1} + 2^{2^0 + 2^1 + ... + 2^{3-n}}x_1 \ge 0 + 2^{2^0 + 2^1 + ... + 2^{2-n}}x_0$, as desired.

This solution was posted and copyrighted by tastymath75025. The original thread for this problem can be found here: [2]

See Also

1982 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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