<s>[INST] 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?
 [/INST] 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.
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