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!
 A: $f(x)=\sqrt{x}$ 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 $\mathbb{R}^+$.
A: 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 $\;65>45$.
A: Note that $$\sqrt { 5 } -\sqrt { 3 } =\frac { \left( \sqrt { 5 } -\sqrt { 3 }  \right) \left( \sqrt { 5 } +\sqrt { 3 }  \right)  }{ \sqrt { 5 } +\sqrt { 3 }  } =\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 }  } ,\\ \quad \sqrt { 15 } -\sqrt { 13 } =\frac { \left( \sqrt { 15 } -\sqrt { 13 }  \right) \left( \sqrt { 15 } +\sqrt { 13 }  \right)  }{ \sqrt { 15 } +\sqrt { 13 }  } =\frac { 2 }{ \sqrt { 15 } +\sqrt { 13 }  } \\ $$
$$\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 }  } >\quad \frac { 2 }{ \sqrt { 15 } +\sqrt { 13 }  } $$
A: $$\sqrt{5}-\sqrt{3}>\sqrt{15}-\sqrt{13}$$
multiply by $(\sqrt{5}+\sqrt{3})(\sqrt{15}+\sqrt{13})$
$$(5-3)(\sqrt{15}+\sqrt{13})>(15-13)(\sqrt{5}+\sqrt{3})$$
or
$$\sqrt{15}+\sqrt{13}>\sqrt{5}+\sqrt{3}$$
A: $\displaystyle{\,\sqrt{\,{a + 2}\,}\, - \,\sqrt{\,{a}\,}\, =
{2 \over \,\sqrt{\,{a + 2}\,}\, + \,\sqrt{\,{a}\,}\,}}$ is a '$\underline{decreasing}$' function of $\displaystyle{a \geq 0}$. Namely,
$$
\,\sqrt{\,{3 + 2}\,}\, - \,\sqrt{\,{3}\,}\, >
\,\sqrt{\,{13 + 2}\,}\, - \,\sqrt{\,{13}\,}\,
$$
A: We need to prove that $\sqrt{13}+\sqrt{5}>\sqrt{15}+\sqrt{3}$, which is Karamata 
because $f(x)=\sqrt{x}$ is concave function and $(15,3)\succ(13,5)$.
A: \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}
