Giải thích các bước giải:
f.$A=1+\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+..++\dfrac{1}{\sqrt{n}}$
$\rightarrow A=\dfrac{2}{2}+\dfrac{2}{2\sqrt{1}}+\dfrac{2}{2\sqrt{2}}+..++\dfrac{2}{2\sqrt{n}}$
$\rightarrow A>\dfrac{2}{1+\sqrt{2}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}+...+\dfrac{2}{\sqrt{n}+\sqrt{n+1}}$
$\rightarrow A>\dfrac{2(\sqrt{2}-1)}{(\sqrt{2}-1)(\sqrt{2}+1)}+\dfrac{2(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}+...+\dfrac{2(\sqrt{n+1}-\sqrt{n})}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}$
$\rightarrow A>\dfrac{2(\sqrt{2}-1)}{2-1}+\dfrac{2(\sqrt{3}-\sqrt{2})}{3-2}+...+\dfrac{2(\sqrt{n+1}-\sqrt{n})}{n+1-n}$
$\rightarrow A>2(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+...+2(\sqrt{n+1}-\sqrt{n})$
$\rightarrow A>2\sqrt{n+1}-2$
$\rightarrow A>\sqrt{n+1}+\sqrt{n+1}-2>\sqrt{n}$
