Sum of irrationals in disguise

Assume $\alpha\in\mathbb{Q}$. $\alpha=\sqrt2+\sqrt3+\sqrt5$.

Square. $\alpha^2=10+2\sqrt6+2\sqrt{10}+2\sqrt{15}$, so $\sqrt6+\sqrt{10}+\sqrt{15}\in\mathbb{Q}$.

Square again. $(\sqrt6+\sqrt{10}+\sqrt{15})^2=31+2(\sqrt{60}+\sqrt{90}+\sqrt{150})=31+2(2\sqrt{15}+3\sqrt{10}+5\sqrt6)$.

Iterate. This gives a new linear combination of $\sqrt6,\sqrt{10},\sqrt{15}$ in $\mathbb{Q}$, forcing a linear relation with rational coefficients. Combining with the first, one finds $\sqrt6\in\mathbb{Q}$, contradiction.

Conclusion. $\alpha$ is irrational.