Lời giải:
Ta thấy:
\(\frac{1}{2}\text{VP}=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{100}}\)
\(> \frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{100}+\sqrt{101}}\)
Mà:
\(\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{100}+\sqrt{101}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3}(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3)}}+...+\frac{\sqrt{101}-\sqrt{100}}{(\sqrt{100}+\sqrt{101})(\sqrt{101}-\sqrt{100})}\)
\(=\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{101}-\sqrt{100}}{101-100}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{101}-\sqrt{100}\)
\(=\sqrt{101}-\sqrt{2}\)
Do đó: \(\frac{1}{2}\text{VP}> \sqrt{101}-\sqrt{2}\Rightarrow \text{VP}>2(\sqrt{101}-\sqrt{2})> 17\) (đpcm)
