Question

chứng minh

\(\dfrac{1}{\left(n+1\right).\sqrt{n}}< \dfrac{2}{\sqrt{n}}-\dfrac{2}{\sqrt{n}+1}\)

với n >1 và n∈N

Answer 1

Lời giải:

Ta có:

Với mọi \(n\in\mathbb{N}>1\Rightarrow 2n+1-\sqrt{n}=(\sqrt{n}-1)^2+n+\sqrt{n}>0\)

\(\Rightarrow 2n+1>\sqrt{n}\)

\(\Rightarrow 2(n+1)> \sqrt{n}+1\)

\(\Rightarrow 2(n+1)\sqrt{n}>(\sqrt{n}+1)\sqrt{n}\)

\(\Rightarrow \frac{1}{2(n+1)\sqrt{n}}< \frac{1}{\sqrt{n}(\sqrt{n}+1)}=\frac{\sqrt{n}+1-\sqrt{n}}{\sqrt{n}(\sqrt{n}+1)}\)

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

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