Prove that using proof by contradiction. √2 + √6 < √15

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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 $\sqrt{6}+\sqrt{2}<\sqrt{15}.$