Easy mistake in proof by contradiction

Suppose we have to prove that $\sqrt{2}+\sqrt{6} < \sqrt{15}$. Now most people would do this way $(\sqrt{2}+\sqrt{6})^2 < (\sqrt{15})^2$ $\Rightarrow 2+6+2\sqrt{2}\sqrt{6} <15$ $\Rightarrow 2\sqrt{12} <7$ $\Rightarrow 48 <49$ However this is not the correct proof because $p \Rightarrow q$ is true even if $latex p$ is false. Therefore the correct way to do this will be let $\sqrt{2}+\sqrt{6} \ge \sqrt{15}$ $\Rightarrow (\sqrt{2}+\sqrt{6})^2 \ge (\sqrt{15})^2$ $\Rightarrow 2+6+2\sqrt{2}\sqrt{6} \ge 15$ $\Rightarrow 2\sqrt{12} \ge 7$ $\Rightarrow 48 \ge 49$ which is false. Hence we proved it by method of proof by contradiction