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?
 A: You can easily show (probably you have already done so) $\sqrt{5},\sqrt{7} \in \mathbb Q(\sqrt{5}+\sqrt{7})$.
If your sum would be rational, we would deduce $\sqrt{5},\sqrt{7} \in \mathbb Q(\sqrt{5}+\sqrt{7}) \subset \mathbb Q(\sqrt{3})$, clearly a contradiction.
A: 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 $.
A: Assume $$\sqrt{3}+ \sqrt{5}+ \sqrt{7}= \frac{a}{b}$$ for some integers $a,b$. Multiblying both sides by $\sqrt{3}+ \sqrt{5}- \sqrt{7}$ yields $$ (\sqrt{3}+\sqrt{5})^2 - 7 = \frac{a(\sqrt{3}+ \sqrt{5}- \sqrt{7})}{b}=\frac{a\left(a/b-2\sqrt{7}\right)}{b} \\ 2\sqrt{15}+8-7=\frac{a^2/b-2a\sqrt{7}}{b} \\ 2b\sqrt{15} = \frac{a^2}{b}-2a\sqrt{7}-b \\ 60b^2=\frac{a^4}{b^2}+28a^2+b^2-\frac{4a^3\sqrt{7}}{b}+4ab\sqrt{7}-2a^2 \\ 59b^2-26a^2-\frac{a^4}{b^2}=4a\sqrt{7}\left(b-\frac{a^2}{b}\right). $$ We can divide both sides by $4a(b^2−a^2)=4a(b-a)(b+a)$ because $a$ can't be $0$ by definition, $b+a$ is a positive integer and $b−a$ can't be $0$ because that would imply $a/b=1$, which is impossible. Thus we get $$\frac{b\left(59b^4-26a^2b^2-a^4\right)}{4a(b^2-a^2)}=\sqrt{7},$$ contradiction.
A: 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.
A: According to Galois, if $\sqrt3 +\sqrt5+\sqrt7$ is rational, then it must be invariant after changing the sign of any or all of the square roots. But the expression is clearly positive, so it cannot be equal to its negative.
