Prove by contradiction (not using a calculator) that $\sqrt6 + \sqrt2 < \sqrt{15}$.
How do you approach such a problem? I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. I tried to square it as I read online but I'm stuck with having $\sqrt3$ and not knowing how to get rid of it to have the answer as clear as the sun.
squaring them yielded:
$(\sqrt2 + \sqrt6)^2 = 8 + 4\sqrt3 < 15$ which indeed holds but the problem is that I am not allowed to use calculator to check the little difference.
My final answer was $4\sqrt3 < 7$.
Best Answer
Let$$\sqrt6+\sqrt2\geq\sqrt{15}.$$ Thus, $$8+2\sqrt{12}\geq15$$ or $$4\sqrt3\geq7$$ or $$48\geq49,$$ which is contradiction.
Id est, our assuming was wrong, which says $$\sqrt6+\sqrt2<\sqrt{15}.$$