Ta có:
\(\frac{1}{\sqrt{1}+\sqrt{2}}>\frac{1}{\sqrt{2}+\sqrt{3}};\frac{1}{\sqrt{3}+\sqrt{4}}>\frac{1}{\sqrt{4}+\sqrt{5}};...;\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{\sqrt{80}+\sqrt{81}}\)
Do đó \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{79}+\sqrt{80}}\right)\)\(>\frac{1}{2}\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
\(=\frac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{80}-\sqrt{79}+\sqrt{81}-\sqrt{80}\right)\)
\(=\frac{1}{2}\left(-\sqrt{1}+\sqrt{81}\right)=\frac{1}{2}\left(-1+9\right)=4\)
Suy ra đpcm.
Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{80}+\sqrt{79}}\)
Suy ra
\(2A=2\left(\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\right)\)
\(=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)
\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+....+\left(\sqrt{80}-\sqrt{79}\right)+\left(\sqrt{81}-\sqrt{79}\right)\)
\(=\sqrt{81}-1=9-1=8\Rightarrow2A>8\Leftrightarrow A>8\)( Đpcm)
