Giả sử \(8>\sqrt{15}+\sqrt{17}\)
\(\Leftrightarrow64>32+2\sqrt{15×17}\)
\(\Leftrightarrow16>\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\left(dung\right)\)
Vậy \(8>\sqrt{15}+\sqrt{17}\)
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