Đặt \(A=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{225}}\)
\(\Leftrightarrow A=\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{3}}+...+\dfrac{2}{\sqrt{225}+\sqrt{225}}\)
\(\Rightarrow A< \dfrac{2}{\sqrt{2}+\sqrt{1}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}+...+\dfrac{2}{\sqrt{225}+\sqrt{224}}=\)
\(=2[\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+(\sqrt{225}-\sqrt{224})]\)
\(\Leftrightarrow A< 2.\left(\sqrt{225}-1\right)=2.14=28\left(đpcm\right)\)
Bài toán tổng quát:Chứng minh BĐT sau với \(n\in N;n\ge2\)
\(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
