Inspired by Nihar

$(a-b)^{2}=a^{2}+b^{2}-2ab$

$(a-b)^{2}=4ab-2ab$

$(a-b)^{2}=2ab$

$(a-b)=\sqrt{2ab}$

$\large \frac{\sqrt{a}}{\sqrt{b}} - \frac{\sqrt{b}}{\sqrt{a}}=\sqrt{2}=\boxed{1.414}$

As $a>b>0$ then $4ab\neq 0$ . Thus we can divide throughout by $4ab$ and get $\frac{a^2+b^2}{4ab}=1$ . This gives $\frac{a^2+b^2}{ab}=4$ and $\frac{a}{b}+\frac{b}{a}=4$

$\frac{a}{b}-2+\frac{b}{a}=2$

${ (\frac { \sqrt { a } }{ \sqrt { b } } -\frac { \sqrt { b } }{ \sqrt { a } } ) }^{ 2 }=2$

${ \frac { \sqrt { a } }{ \sqrt { b } } -\frac { \sqrt { b } }{ \sqrt { a } } }=\pm \sqrt { 2 }$

But as $a>b$ , $\frac { \sqrt { a } }{ \sqrt { b } } >\frac { \sqrt { b } }{ \sqrt { a } }$ and thus ${ \frac { \sqrt { a } }{ \sqrt { b } } -\frac { \sqrt { b } }{ \sqrt { a } } }=\sqrt { 2 } { \approx 1.414 }$

$\Rightarrow a^2+b^2=4ab$

$(a-b)^2+2ab=4ab$

$(a-b)^2=2ab$

Wee need to find:

$\dfrac{\sqrt{a}}{\sqrt{b}} - \dfrac{\sqrt{b}}{\sqrt{a}}=\dfrac{a-b}{\sqrt{ab}}=x$

Squring both sides.

$\dfrac{(a-b)^2}{ab}=x^2$

$\dfrac{2ab}{ab}=x^2$

$\therefore x=\sqrt{2}\approx{\boxed{1.414}}$

Let $\dfrac{b}{a} = x$
$1 + x^{2} = 4x$
$\therefore x = 2 \pm \sqrt{3}$
But, since $a \ge b , x < 1 \to x = 2 - \sqrt{3}$
$\sqrt{\dfrac{a}{b}} - \sqrt{\dfrac{b}{a}} = \sqrt{2+\sqrt{3}} - \sqrt{2-\sqrt{3}}= \dfrac{1}{\sqrt{2}} \cdot \left( \sqrt{(\sqrt{3}+1)^{2}} - \sqrt{(\sqrt{3}-1)^{2}}\right) = \sqrt{2}$

Given that,

$a^{2} + b^{2} = 4ab$

$\implies a^{2} -2ab + b^{2} = 2ab$

$\implies (a-b)^{2} = 2ab$

$\implies (a-b) = \sqrt{2ab}$

Now, $\frac{\sqrt{a}}{\sqrt{b}} - \frac{\sqrt{b}}{\sqrt{a}} = \frac{a-b}{\sqrt{ab}} = \frac{\sqrt{2ab}}{\sqrt{ab}} = \frac{\sqrt{2} \times \sqrt{ab}}{\sqrt{ab}} = \sqrt{2} ≈ \boxed{1.414}$