Prove by contradiction (not using a calculator) that $\sqrt6 + \sqrt2 < \sqrt{15}$? 
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 $4\sqrt3 < 7$.
 A: 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
$$\sqrt6+\sqrt2<\sqrt{15}.$$
A: The claim is that $\sqrt{2} + \sqrt{6} < \sqrt{15}$, so its negation is simply the statement

$\sqrt{2} + \sqrt{6} < \sqrt{15}$ is false

or alternatively,

$\sqrt{2} + \sqrt{6} \ge \sqrt{15}$

Now in order to prove the claim, suppose that $\sqrt{2} + \sqrt{6} \ge \sqrt{15}$ and square both sides, getting
$$2 + 6 + 2\sqrt{12} \ge 15$$
Simplifying, this leads to
$$2 \sqrt{12} \ge 7$$
Do you see how to derive a contradiction now?
A: 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
$$\sqrt6+\sqrt2<\sqrt{15}.$$
