<s>[INST] Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational. Note that my attempted method below is distinct from the solutions in this question.
I also know this is generally true for $\sqrt p + \sqrt q + \sqrt r$ where $p,q,r$ are prime, but I am asking if my particular method works.
Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.
Assume that $\sqrt 2 + \sqrt 3 + \sqrt 5 = \frac pq$ for some $p,q \in \mathbb Z$, in lowest terms, with $q \neq 0$. Then at most one of $p$ and $q$ can be even.  Then
\begin{align*}
\sqrt 2 + \sqrt 3 + \sqrt 5 &= \frac pq\\
\\
(\sqrt 2 + \sqrt 3 + \sqrt 5)^2 &= \left(\frac pq \right)^2\\
\\
2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=\frac{p^2}{q^2}\\
\\
2q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=p^2
\end{align*}
which shows that $p^2$ is even, so $p$ is even and let $p=2k$ for some $k\in \mathbb Z$. Substituting this we get
\begin{align*}
2q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=(2k)^2\\
\\
q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=2k^2\\
\\
q^2 &= 2 \left(\frac{k^2}{5+\sqrt 6 + \sqrt{10} + \sqrt{15}}\right),
\end{align*}
which shows as well that $q$ is even, a contradiction. Hence $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.
 [/INST] 
$$2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})=\frac{p^2}{q^2}$$

This step, you cannot conclude the LHS is an even integer.
You can try this: Assume $\sqrt{2}+\sqrt{3}+\sqrt{5}=r_1$ be rational.
$$\sqrt{2}+\sqrt{3}=r_1-\sqrt{5}\Leftrightarrow 2\sqrt{6}=r_1^2-2\sqrt{5}r_1\Leftrightarrow \sqrt{6}+\sqrt{5}r_1=\frac{r_1^2}{2}=r_2 \\
\\
\Leftrightarrow \sqrt{6}=r_2-\sqrt{5}r_1\Leftrightarrow 6=r_2^2+5r_1^2-2r_1\sqrt{5}\Leftrightarrow \sqrt{5}=\frac{r_2^2+5r_1^2-6}{2r_1}$$
$\sqrt{5}$ is irrational , but RHS is rational, so you get contradictions.
</s>