Chủ đề này có 1 trả lời

[nguyenducthanh]

Cho A = $A=\frac{\sqrt{2}-\sqrt{1}}{2+1}+\frac{\sqrt{3}-\sqrt{2}}{3+2}+\frac{\sqrt{4}-\sqrt{3}}{4+3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1+n} CMR:A <\frac{\sqrt{n+1}-1}{2\sqrt{n+1}}$

[Hero Crab]

Áp dụng bđt AM-GM cho 2 số dương ta có

$n+1+n>2 \sqrt{n\left ( n +1\right )}$

$\frac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left ( n+1 \right )}}=\frac{1}{2}.\left (\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}  \right )$

$\Rightarrow \frac{\sqrt{2}-1}{2+1}<\frac{1}{2}\left (1- \frac{1}{\sqrt{2}} \right )$

$\frac{\sqrt{3}-\sqrt{2}}{3+2}<\frac{1}{2}\left (\frac{1}{\sqrt{2}}- \frac{1}{\sqrt{3}} \right )$

$\frac{\sqrt{4}-\sqrt{3}}{4+3}<\frac{1}{2}\left (\frac{1}{\sqrt{3}}- \frac{1}{\sqrt{4}} \right )$

Cộng theo vế lại $\Rightarrow \frac{\sqrt{2}-1}{2+1}+\frac{\sqrt{3}-\sqrt{2}}{3+2}+\frac{\sqrt{4}-\sqrt{3}}{4+3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1+n}<\frac{1}{2}.\left ( 1-\frac{1}{\sqrt{n+1}} \right )=\frac{\sqrt{n+1}-1}{2\sqrt{n+1}}$ (ĐPCM)