Prove that $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Generalise this.

Hint:

Assume by contradiction that $\psi=\sqrt{2}+\sqrt{3}+\sqrt{5}$ is rational. Then $$(\psi-\sqrt{5})^2=(\sqrt{2}+\sqrt{3})^2 \\ \psi^2-2\sqrt{5}\psi+5=5+2\sqrt{6} \\ \psi^2=2\sqrt{6} +2\sqrt{5}\psi\\ $$

Square it one more time, and you reach the contradiction.

For the general case, here is a nice trivial solution from Kvant: prove the following lemma:

Lemma If $a_1,a_2,..,a_k$ are distinct integers $\geq 2$, none of which is divisible by a square, and $b_1,..,b_n$ are integers such that $$b_1\sqrt{a_1}+...+b_n\sqrt{a_n} \in \mathbb Q$$ then $b_1=...=b_n=0$.

Proof: Do induction by the number $m$ of primes which divide $a_1...a_n$.

The inductive step $P(m)\Rightarrow P(m+1)$ is done by contradiction: move all terms for which $a_k$ is divisible by the $p_{m+1}$ on one side, everything else on the other side and square, to end in the case $P_m$.

Okay, if $\sqrt{2}+\sqrt{3}+\sqrt{5}=r\in\mathbb Q$, then: $$(\sqrt{2}+\sqrt{3})^2=(r-\sqrt{5})^2$$ $$2+2\sqrt6+3=r^2-2\sqrt5 r+5$$ $$2\sqrt6=r^2-2\sqrt5 r$$ Square this once again and you obtain that $\sqrt5$ is rational, which is a contradiction.

We assume we have $$\sqrt{2}+\sqrt{3}+\sqrt{5}=\frac{m}{n}$$ with $\gcd(m,n)=1$. We write $$\sqrt{2}+\sqrt{3}=\frac{m}{n}-\sqrt{5};$$ squaring, we have $$2\sqrt{6}=\frac{m^2}{n^2}-2\frac{m}{n}\sqrt{5};$$ and squaring once again then simplifying, we arrive at $$5\frac{n}{m}+\frac{m}{4n}-6\frac{n^3}{m^3}=\sqrt{5}.$$

This is a contradiction, since we have on the left only rational numbers and on the right side an irrational number.