1969 Canadian MO Problems/Problem 2

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


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

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