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

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

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?