Question

Chứng minh : Với mọi số tự nhiên n , ta có : \(\sqrt{n+1}-\sqrt{n}>\dfrac{1}{2\sqrt{n+1}}\)

Answer 1

ta có: \(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\)

CM: ta có \(VP=\dfrac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)

\(VP=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}=VT\)

ta có bpt \(\Leftrightarrow\dfrac{1}{\sqrt{n+1}+\sqrt{n}}>\dfrac{1}{2\sqrt{n+1}}\)

\(\Leftrightarrow\sqrt{n+1}+\sqrt{n}< 2\sqrt{n+1}\)

mà \(n< n+1\Rightarrow\sqrt{n}< \sqrt{n+1}\)

\(\Rightarrow\sqrt{n+1}+\sqrt{n}< 2\sqrt{n+1}\) (đpcm)