Giải:
Ta có:
\(\sqrt{1}< \sqrt{n}\Leftrightarrow\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{n}}\)
\(\sqrt{2}< \sqrt{n}\Leftrightarrow\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{n}}\)
\(\sqrt{3}< \sqrt{n}\Leftrightarrow\dfrac{1}{\sqrt{3}}>\dfrac{1}{\sqrt{n}}\)
...
\(\sqrt{n}=\sqrt{n}\Leftrightarrow\dfrac{1}{\sqrt{n}}=\dfrac{1}{\sqrt{n}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}>\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n}}+...+\dfrac{1}{\sqrt{n}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}>\dfrac{n}{\sqrt{n}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}>\sqrt{n}\)
Vậy ...
