Answer
$\frac{5\sqrt{2}+3 \sqrt{3}-4 \sqrt{6}+2}{23}$
Work Step by Step
Do the rationalization as shown below.
\begin{equation}
\begin{aligned}
\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{4}} & = \frac{1}{2+\left(\sqrt{2}+\sqrt{3}\right)} \\
& =\frac{1}{2+\left(\sqrt{2}+\sqrt{3}\right)}\cdot \frac{2-\left(\sqrt{2}+\sqrt{3}\right)}{2-\left(\sqrt{2}+\sqrt{3}\right)} \\
& = \frac{2-\left(\sqrt{2}+\sqrt{3}\right)}{4-\left(\sqrt{2}+\sqrt{3}\right)^2}\\
&=\frac{2-(\sqrt{2}+\sqrt{3})}{-(1+2 \sqrt{6})}\\
&=\frac{(\sqrt{2}+\sqrt{3})-2}{(1+2 \sqrt{6})}\cdot \frac{\left(1-2\sqrt{6}\right)}{\left(1-2\sqrt{6}\right)} \\
&=\frac{\sqrt{2}(1-2 \sqrt{6})+\sqrt{3}(1-2 \sqrt{6})-2(1-2 \sqrt{6})}{1-4(6)}\\
&= \frac{\sqrt{2}-2\sqrt{12}+\sqrt{3}-2\sqrt{18}-2+4\sqrt{6}}{-23}\\
&= -\frac{\sqrt{2}-4\sqrt{3}+\sqrt{3}-6 \sqrt{2}-2+4 \sqrt{6}}{23}\\
& = -\frac{-5\sqrt{2}-3 \sqrt{3}+4 \sqrt{6}-2}{23}\\
&= \frac{5\sqrt{2}+3 \sqrt{3}-4 \sqrt{6}+2}{23}
\end{aligned}
\end{equation}
where
$$ \left(\sqrt{2}+\sqrt{3}\right)^2= 2+2\sqrt{2}\cdot \sqrt{3}+3= 5+2\sqrt{6} $$
and
$$ 4-\left(\sqrt{2}+\sqrt{3}\right)^2= 4-5-2\sqrt{6}= -1-2\sqrt{6}$$
The solution is:
$$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{4}} = \frac{5\sqrt{2}+3 \sqrt{3}-4 \sqrt{6}+2}{23}$$
