Prove by contradiction (not using a calculator) that $\sqrt6 + \sqrt2 < \sqrt{15}$?

Question

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 $4\sqrt3 < 7$.

Answer 1

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 $$\sqrt6+\sqrt2<\sqrt{15}.$$

Answer 2

This is an old question but recently mentioned. I wanted to comment that there is no gain from using contradiction:

A proof by contradiction is like

claim $2*3=6$. Proof: assume otherwise. But $2*3=3+3=6. $CONTRADICTION TO WHAT WE ASSUMED!!!!

Here is the first proof reconfigured. I tried to stick to the form there.

Claim: $$\sqrt6+\sqrt2\lt \sqrt{15}.$$ Proof: Equivalently (squaring both sides) the claim is $$8+2\sqrt{12}\lt 15$$ or $$4\sqrt3\lt 7$$ or (squaring again) $$48\lt 49,$$ which is true.

Id est, our claim is true , which says $$\sqrt6+\sqrt2<\sqrt{15}.$$