Prove that $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Generalise this. I'm reading R. Courant & H. Robbins' "What is Mathematics: An Elementary Approach to Ideas and Methods" for fun. I'm on page $60$ and $61$ of the second edition. There are three exercises on proving numbers irrational spanning these pages, the last is as follows.

Exercise $3$: Prove that $\phi=\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Try to make up similar and more general examples.

My Attempt:

Lemma: The number $\sqrt{2}+\sqrt{3}$ is irrational. (This is part of Exercise 2.)

Proof: Suppose $\sqrt{2}+\sqrt{3}=r$ is rational. Then
$$\begin{align}
2&=(r-\sqrt{3})^2 \\
&=r^2-2r\sqrt{3}+3
\end{align}$$ is rational, so that
$$\sqrt{3}=\frac{r^2+1}{2r}$$ is rational, a contradiction. $\square$
Let $\psi=\sqrt{2}+\sqrt{3}$. Then, considering $\phi$,
$$\begin{align}
5&=(\phi-\psi)^2 \\
&=\phi^2-2\psi\phi+5+2\sqrt{6}.
\end{align}$$
I don't know what else to do from here.
My plan is/was to use the Lemma above as the focus for a contradiction, showing $\psi$ is rational somehow.
Please help :)
Thoughts:
The "try to make up similar and more general examples" bit is a little vague.

The question is not answered here as far as I can tell.

Using the straightforward fact that the sum of two (or more) algebraic integers is an algebraic integer (see here), one has the following very short argument: since $\sqrt{2},\sqrt{3},\sqrt{5}$ are algebraic integers, their sum $A=\sqrt{2}+\sqrt{3}+\sqrt{5}$ must be also. Hence if $A$ is rational, it must be an integer. However, as a rudimentary estimate of the three square roots shows, we have $5 < A < 6$: contradiction.
