Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational 
How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?

I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?

Suppose $\sqrt{3}+\sqrt{5}+\sqrt{7} = r$ for some rational $r$. Then, $$(\sqrt{3}+\sqrt{5})^2 = (r-\sqrt{7})^2 \implies 8+2\sqrt{15} = 7+r^2-2r\sqrt{7}$$
So, $$1-r^2+2\sqrt{15} =-2r\sqrt{7}$$
Let $1-r^2 = k$, which will be a rational number. So,
$$(k+2\sqrt{15})^2 = k^2+ 60+4k\sqrt{15} = 28r^2$$
The LHS is irrational while the RHS is rational. Hence, we have a contradiction.
