How to prove $\sqrt{5}-\sqrt{3}$ is bigger than $\sqrt{15}-\sqrt{13}$

Note that $$\sqrt { 5 } -\sqrt { 3 } =\frac { \left( \sqrt { 5 } -\sqrt { 3 } \right) \left( \sqrt { 5 } +\sqrt { 3 } \right) }{ \sqrt { 5 } +\sqrt { 3 } } =\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 } } ,\\ \quad \sqrt { 15 } -\sqrt { 13 } =\frac { \left( \sqrt { 15 } -\sqrt { 13 } \right) \left( \sqrt { 15 } +\sqrt { 13 } \right) }{ \sqrt { 15 } +\sqrt { 13 } } =\frac { 2 }{ \sqrt { 15 } +\sqrt { 13 } } \\ $$

$$\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 } } >\quad \frac { 2 }{ \sqrt { 15 } +\sqrt { 13 } } $$

$f(x)=\sqrt{x}$ is a concave function on $\mathbb{R}^+$, hence for any fixed $h>0$ the function $$\Delta_h(x) = f(x+h)-f(x) $$ is a decreasing function on $\mathbb{R}^+$.

It amounts to proving $\;\sqrt 5+\sqrt{13}>\sqrt 3+\sqrt{15}$.

As everyone is positive, you can compare the squares: $$(\sqrt 5+\sqrt{13})^2=18+2\sqrt{65}\quad\text{vs}\quad (\sqrt 3+\sqrt{15})^2=18+2\sqrt{45}.$$ Indeed $\;65>45$.