Prove that using proof by contradiction.
√2 + √6 < √15
Assume the result is False; i.e. assume that 6+2≥15.\sqrt{6}+\sqrt{2}\geq\sqrt{15}.6+2≥15.
Then
which is contradiction.
Hence we have proved that 6+2<15.\sqrt{6}+\sqrt{2}<\sqrt{15}.6+2<15.
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