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

Question

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!

Answer 1

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

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

Answer 2

$f(x)=\sqrt{x}$ 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 $\mathbb{R}^+$.

Answer 3

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 $\;65>45$.

Answer 4

$$\sqrt{5}-\sqrt{3}>\sqrt{15}-\sqrt{13}$$ multiply by $(\sqrt{5}+\sqrt{3})(\sqrt{15}+\sqrt{13})$ $$(5-3)(\sqrt{15}+\sqrt{13})>(15-13)(\sqrt{5}+\sqrt{3})$$ or $$\sqrt{15}+\sqrt{13}>\sqrt{5}+\sqrt{3}$$

Answer 5

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 $(15,3)\succ(13,5)$.

Answer 6

$\displaystyle{\,\sqrt{\,{a + 2}\,}\, - \,\sqrt{\,{a}\,}\, = {2 \over \,\sqrt{\,{a + 2}\,}\, + \,\sqrt{\,{a}\,}\,}}$ is a '$\underline{decreasing}$' function of $\displaystyle{a \geq 0}$. Namely, $$ \,\sqrt{\,{3 + 2}\,}\, - \,\sqrt{\,{3}\,}\, > \,\sqrt{\,{13 + 2}\,}\, - \,\sqrt{\,{13}\,}\, $$

Answer 7

\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}