Answer to Question #255912 in Discrete Mathematics for htd

Question

Prove that using proof by contradiction.

√2 + √6 < √15

Answer 1

Assume the result is False; i.e. assume that $\sqrt{6}+\sqrt{2}\geq\sqrt{15}.$

Then

$$\sqrt{6}+\sqrt{2}\geq\sqrt{15}>0=>(\sqrt{6}+\sqrt{2})^2\geq(\sqrt{15})^2$$

$$=>6+2\sqrt{12}+2\geq15=>4\sqrt{3}\geq7$$

$$=>(4\sqrt{3})^2\geq(7)^2=>48\geq49,$$

which is contradiction.

Hence we have proved that $\sqrt{6}+\sqrt{2}<\sqrt{15}.$