# Difference between revisions of "1969 Canadian MO Problems/Problem 2"

## Problem

Determine which of the two numbers $\sqrt{c+1}-\sqrt{c}$, $\sqrt{c}-\sqrt{c-1}$ is greater for any $c\ge 1$.

## Solution

Multiplying and dividing $\sqrt{c+1}-\sqrt c$ by its conjugate,

$\sqrt{c+1}-\sqrt c=\frac{(\sqrt{c+1})^2-(\sqrt c)^2}{\sqrt{c+1}+\sqrt{c}}=\frac1{\sqrt{c+1}+\sqrt{c}}.$

Similarly, $\sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c-\sqrt{c-1}}$. We know that $\frac1{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt c-\sqrt{c-1}}$ for all positive $c$, so $\sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}$.

 1969 Canadian MO (Problems) Preceded byProblem 1 1 • 2 • 3 • 4 • 5 Followed byProblem 3