1969 Canadian MO Problems/Problem 2


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.


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

Invalid username
Login to AoPS