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 1

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 $\sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}$ .

Solution 2

Considering the derivative of $f(x)=\sqrt{x+1}-\sqrt{x}$ .

We have $f'(x)=\frac{1}{2\sqrt{x+1}}-\frac{1}{2\sqrt{x}}$ . Putting under a common denominator, we can see that the top will be negative.

Thus $\boxed{\sqrt{c}-\sqrt{c-1}}$ is greater.

~hastapasta

Solution 3

Plugging in $1$ for both of the expressions, we get that $\sqrt{c+1} + \sqrt{c} = \sqrt{2} - \sqrt{1} = \sqrt{2} - 1$ and $\sqrt{c} - \sqrt{c-1} = \sqrt{1} - \sqrt{0} = 1$ . Since $\sqrt{2} - 1 < 1$ , $\boxed{\sqrt{c} - \sqrt{c-1}}$ is greater -andliu766

1969 Canadian MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •	Followed by Problem 3

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1969 Canadian MO Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

Solution 3