Proving the Equality of Two Fractions Using Algebra

[Kara386]

Homework Statement

This is something I need to show in order to solve the question I've been asked. I need to show that
$\sqrt{\frac{N_A}{N_D} } + \sqrt{\frac{N_D}{N_A} } = \frac{N_A + N_B}{\sqrt{N_A N_B}}$
I know these two sides are equal because wolfram alpha says they are, and also because it works if I sub that into my proof. But I'm doing something really really really stupid I think, because I can't get there.

Homework Equations

The Attempt at a Solution

I thought it might be easiest to square this and simplify, then square root. So
$(\sqrt{\frac{N_A}{N_D} } + \sqrt{\frac{N_D}{N_A} })^2 = \frac{N_A}{N_D} + 2\sqrt{\frac{N_A N_B}{N_A N_B}} + \frac{N_D}{N_A}$
I suspect that's the step that's wrong but I don't know why. Carrying on anyway:
$= \frac{N_A}{N_D} + \frac{N_D}{N_A} + 2$
Finding a common denominator:

$= \frac{N_A N_D + N_D N_A + 2N_A N_D}{N_A N_D}$
Then finally square rooting again:
$=2$
So that's very different to what I'm after and there is some really awful algebra mistake in there. Unfortunately I can't find it, any help would be very much appreciated! :)

 

[Kara386]

Oh yes, that is me being pretty stupid. Could just multiply first term by $\frac{N_A}{N_A}$ and the second term by $\frac{N_D}{N_D}$, then:
$\sqrt{\frac{N_A^2}{N_A N_D}} + \sqrt{\frac{N_D^2}{N_A N_D} }= \frac{N_A}{\sqrt{N_A N_D}} + \frac{N_D}{\sqrt{N_A N_D}}$

 

[BvU]

Finding a common denominator

leads to $N_A^2 + 2N_A N_D + N_D^2 \over N_A N_D$

 

[Kara386]

leads to $N_A^2 + 2N_A N_D + N_D^2 \over N_A N_D$

Ah, that too! Oops! Thank you, I was never going to get that.