Question

Which is greater $\sqrt{7}-\sqrt{3}$ or $\sqrt{5}-1$?

Answer 1

Let $x=\sqrt{7}-\sqrt{3}\&y=\sqrt{5}-1$

$\frac{1}{x}=\frac{1}{\sqrt{7}-\sqrt{3}}\times \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{\sqrt{7}+\sqrt{3}}{7-3}=\frac{\sqrt{7}+\sqrt{3}}{4}$

$\frac{1}{y}=\frac{1}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{5-1}=\frac{\sqrt{5}+1}{4}$

clearly

$\frac{1}{x}>\frac{1}{y}$

$\Rightarrow y>x$

$\Rightarrow \sqrt{5}-1>\sqrt{7}-\sqrt{3}$

Therefore $\sqrt{5}-1$ is greater