<s>[INST] 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!
 [/INST] Note that $$\sqrt { 5 } -\sqrt { 3 } =\frac { \left( \sqrt { 5 } -\sqrt { 3 }  \right) \left( \sqrt { 5 } +\sqrt { 3 }  \right)  }{ \sqrt { 5 } +\sqrt { 3 }  } =\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 }  } ,\\ \quad \sqrt { 15 } -\sqrt { 13 } =\frac { \left( \sqrt { 15 } -\sqrt { 13 }  \right) \left( \sqrt { 15 } +\sqrt { 13 }  \right)  }{ \sqrt { 15 } +\sqrt { 13 }  } =\frac { 2 }{ \sqrt { 15 } +\sqrt { 13 }  } \\ $$
$$\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 }  } >\quad \frac { 2 }{ \sqrt { 15 } +\sqrt { 13 }  } $$
</s>