Rationalizing radicals Rationalize
(i.e., get rid of radicals in the denominator)
in
$\dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$,
$\dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}$,
and,
in general
$\dfrac{1}{\sum_{i=1}^n \sqrt{a_i}}$.
I have submitted my work so far as an answer.

I'll work out the first case,
and, after that,
show my initial work on doing
the general case by induction.
$\begin{align}
\dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}
&= \dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}
\dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}
\\
&=\dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{(\sqrt{a}+\sqrt{b})^2-c}\\
&=\dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{a+b-c+2\sqrt{ab}}\\
&=\dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{a+b-c+2\sqrt{ab}}
\dfrac{a+b-c-2\sqrt{ab}}{a+b-c-2\sqrt{ab}}\\
&=\dfrac{(\sqrt{a}+\sqrt{b}-\sqrt{c})(a+b-c-2\sqrt{ab})}{(a+b-c)^2-4ab}\\
\end{align}
$
Here is an attempt
to do the general case
by induction.
If
$\dfrac{1}{\sum_{i=1}^n \sqrt{a_i}}
=\dfrac{F_n}{G_n}
$,
where $G_n$ is radical-free,
$\begin{align}
\dfrac{1}{\sum_{i=1}^{n+1} \sqrt{a_i}}
&= \dfrac{1}{\sum_{i=1}^n \sqrt{a_i}+\sqrt{a_{n+1}}}\\
&= \dfrac{1}{\sum_{i=1}^n \sqrt{a_i}+\sqrt{a_{n+1}}}
\dfrac{\sum_{i=1}^n \sqrt{a_i}-\sqrt{a_{n+1}}}{\sum_{i=1}^n \sqrt{a_i}-\sqrt{a_{n+1}}}\\
&= \dfrac{\sum_{i=1}^n \sqrt{a_i}-\sqrt{a_{n+1}}}
{(\sum_{i=1}^n \sqrt{a_i})^2-a_{n+1}}\\
&= \dfrac{G_n/F_n-\sqrt{a_{n+1}}}
{(G_n/F_n)^2-a_{n+1}}\\
&= \dfrac{F_nG_n-F_n^2\sqrt{a_{n+1}}}
{G_n^2-F_n^2a_{n+1}}\\
\end{align}
$
I would like to be able to say
at this point that
$F_{n+1} = F_nG_n-F_n^2\sqrt{a_{n+1}}$
and
$G_{n+1} = G_n^2-F_n^2a_{n+1}$.
The problem is that
the denominator is
$not$ radical-free,
since it contains $F_n^2$.
I'll leave it at that for now.
