For example, rationalizing expressions like $$\frac{1}{\pm \sqrt{a} \pm \sqrt{b}}$$ Is straightforward. Moreover cases like $$\frac{1}{\pm \sqrt{a} \pm \sqrt{b} \pm \sqrt{c}}$$ and $$\frac{1}{\pm \sqrt{a} \pm \sqrt{b} \pm \sqrt{c} \pm \sqrt{d}}$$

Are still easy to rationalize. But my question is in the more general case $$\frac{1}{\pm \sqrt{a_1} \pm \sqrt{a_2} \cdots \pm \sqrt{a_n}}$$ Where $n \ge 5$

Are they always rationalizable? If so, how would be an algorithm to rationalize them. If not, then a proof must exist.

From my point of view, I can't find an obvious way to rationalize the case $n=5$, since grouping the radicals in a group of 3 and a group of 2 radicals and then applying the identity $$(a-b)*(a+b)=a^2-b^2$$ Just modifies the denominator from $$\pm \sqrt{a} \pm \sqrt{b} \pm \sqrt{c} \pm \sqrt{d} \pm \sqrt{e}$$ to $$\pm v \pm \sqrt{w} \pm \sqrt{x} \pm \sqrt{y} \pm \sqrt{z}$$ Will this help in something? Or a different method or identity is needed?

Consider $$P_n(t_1,\ldots,t_n) = \prod_{\sigma \in \{-1,1\}^n}\sum_i \sigma_i t_i$$ You probably don't want to expand this explicitly for symbolic $t_i$, but it is a polynomial in the $t_i$ whose terms involve only even powers (because it's invariant under $t_i \to -t_i$). Thus $P_n(\sqrt{a_1},\ldots,\sqrt{a_n})$ is a polynomial in $a_1, \ldots, a_n$. Then for any $\rho \in \{-1,1\}^n$, $$\frac{1}{\sum_i \rho_i \sqrt{a_i}} = \dfrac{\prod_{\sigma \ne \rho} \sum_i \sigma_i \sqrt{a_i}}{P_n(\sqrt{a_1},\ldots,\sqrt{a_n})}$$

Yes, it's always possible to rationalize such expressions.

Without loss of generality, it's enough to consider the first square root having positive sign in front of it. Then, see what happens if you multiply together terms with all possible combinations of the remaining signs:

$$\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)=a-b$$

$$\begin{eqnarray} \left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right) \left(\sqrt{a}+\sqrt{b}-\sqrt{c}\right)\\ \left(\sqrt{a}-\sqrt{b}+\sqrt{c}\right) \left(\sqrt{a}-\sqrt{b}-\sqrt{c}\right) & = & a^2 + b^2 + c^2 - 2ab - 2ac - 2bc \\ \end{eqnarray}$$

The resulting expressions get progressively more complicated (the resulting polynomial will be homogeneous, symmetric polynomial of degree $2^{n-2}$ -- it'll contain all possible combinations of the variables with all possible combinations of exponents), but it won't contain any of the square roots anymore.

The squaring of a sum of $n$ square roots $\pm\sqrt a_i$ produces $n(n-1)/2$ terms of the form $\pm\sqrt {a_ia_j}$. If you partition more than $2n$ terms into two lots, one of the lots has more than $n$ terms. Since $n(n-1)/2>2$ when $n>2$, you can never get below $3$ square-root terms if you start with more than $4$ terms.

Edit: Robert Israel's answer is the way to do it, but this involves more than multiplying $a+b$ by $a-b.$