Ta chứng minh được \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\) với mọi n là số tự nhiên lớn hơn 0
Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
Ta có \(2A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{79}+\sqrt{80}}>\)
\(>\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+...+\sqrt{80}-\sqrt{79}+\sqrt{81}-\sqrt{80}\)
\(=\sqrt{81}-\sqrt{1}=8\)
\(\Rightarrow2A>8\Rightarrow A>4\)
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
\(\rightarrow\)VT = \(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{80}-\sqrt{79}=\sqrt{80}-1>4\)