Motivation
No particular motivation, I'm just curious to see if an interesting result can emerge from such a seemingly simple problem.
Problem
Find the closed symmetric formula for the rationalization of $$\left(\sum_{i=1}^{n}\sqrt{a_i}\right)^{-1}$$
Initial formulas
With $n=1$ is trivial: $$\dfrac{1}{\sqrt{a}}=\dfrac{\sqrt{a}}{a}=\frac{a^{\frac{3}{2}}}{a^2}$$ With $n=2$ we have $$\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{a-b}$$ Doing the average with the version with $a$ and $b$ exchanged we have: $$\dfrac{1}{\sqrt{a}+\sqrt{b}}=\frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}-a\sqrt{b}-\sqrt{a}b}{a^{2}-2ab+b^{2}}$$ With $n=3$ the interesting part begins: rationalizing with the usual conjugate rule you can obtain $3$ different forms (all equivalent to each other): $$\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\begin{cases} \frac{(\sqrt{a}+\sqrt{b}-\sqrt{c})(a+b-c-2\sqrt{ab})}{(a+b-c)^2-4ab}\\ \frac{(\sqrt{b}+\sqrt{c}-\sqrt{a})(b+c-a-2\sqrt{bc})}{(b+c-a)^{2}-4bc}\\ \frac{(\sqrt{c}+\sqrt{a}-\sqrt{b})(c+a-b-2\sqrt{ca})}{(c+a-b)^{2}-4ca}\end{cases}$$ However, these $3$ formulas are not symmetrical, so I averaged them, and I got: $$\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}-a\sqrt{b}-\sqrt{a}b-a\sqrt{c}-\sqrt{a}c-\sqrt{b}c-b\sqrt{c}+2\sqrt{abc}}{a^{2}+b^{2}+c^{2}-2ab-2bc-2ca}$$ For a generic $n$ there will be $n$ different shapes with the "cycled" elements, in the end the symmetric result will be their average.
I would be curious to see if there is a closed formula for generic $n$, I hypothesize that the final formula could be something like:
$$\left(\sum_{i=1}^{n}\sqrt{a_i}\right)^{-1}=\frac{\displaystyle\sum_{i_1+...+i_n=3}\mathcal{C}_{i_1,...,i_n}\sqrt{a_1^{i_1} a_2^{i_2}\cdot...\cdot a_n^{i_n}}}{\displaystyle \sum_{i=1}^{n}a_i^2-2\sum_{1\geq i>j\geq n}a_i a_j}$$ it can be written similarly $$\left(\sum_{i=1}^{n}\sqrt{a_i}\right)^{-1}=\frac{\displaystyle\sum_{|\mathbf{i}|=3}\mathcal{C}_{\mathbf{i}}\sqrt{\mathbf{a}^{\mathbf{i}}}}{\displaystyle \sum_{i=1}^{n}a_i^2-2\sum_{1\geq i>j\geq n}a_i a_j}$$ Where $\mathcal{C}_{i_1,...,i_n}$ is a constant that depends on the partition of $n$. In the second formula I used multi-index notation where $\mathbf{a}=(a_1,...,a_n)$ and $\mathbf{i}=(i_1,...,i_n)$
Generalization
An immediate generalization then would also be to try to find the formula of $$\left(\sum_{i=1}^{n}\sqrt[m]{a_i}\right)^{-1}$$