How to prove $\sqrt{5}-\sqrt{3}$ is bigger than $\sqrt{15}-\sqrt{13}$ Although I can determine using a calculator that $\sqrt{5}-\sqrt{3}$ is larger than $\sqrt{15}-\sqrt{13}$, how would I go about proving that? My teacher gave us a hint which was to use the difference of two squares identity $(a^2-b^2) = (a-b)\cdot(a+b)$, but I don't see how to proceed.
Thanks in advance!

\begin{align}
   (\sqrt 5 - \sqrt 3)(\sqrt 5 + \sqrt 3)
   &= (\sqrt{15} - \sqrt{13})(\sqrt{15} + \sqrt{13}) = 2
\\ 
   \dfrac{\sqrt 5 - \sqrt 3}{\sqrt{15} - \sqrt{13}}
   &= \dfrac{\sqrt{15} + \sqrt{13}}{\sqrt 5 + \sqrt 3} > 1
\\ 
   \sqrt 5 - \sqrt 3 &> \sqrt{15} - \sqrt{13}
\end{align}
