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?

Let $\alpha=\sqrt{3}+ \sqrt{5}+ \sqrt{7}$.
Then $\alpha$ is a root of $x^8-60 x^6+782 x^4-3180 x^2+3481=0$.
Since this is a monic polynomial with integer coefficients, the rational root theorem tells you that $\alpha$ is either irrational or an integer. 
Now
$\quad 1.7 < \sqrt 3 < 1.8 $
$\quad 2.2 < \sqrt 5 < 2.3 $
$\quad 2.6 < \sqrt 7 < 2.7 $
gives
$\quad 6.5 < \alpha < 6.8 $
which proves that $\alpha$ is not an integer.
We can avoid even these fine estimates: if $\alpha$ is an integer then it must divide $3481=59^2$, but clearly $1 < \alpha < 3\sqrt 7 < 9 < 59 $.
