Chứng minh: 1 / (1 + căn 2) + 1 / (căn 3 + căn 4) +...+ 1 / (căn 79 + căn 80) > 4

Đề bài: Chứng minh: $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\mathrm{...}+\frac{1}{\sqrt{79}+\sqrt{80}}>4$ .

Trả lời

Hướng dẫn giải:

Ta có: $\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k}+\sqrt{k+1}}$

$$\begin{gathered} \Rightarrow \frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k}+\sqrt{k+1}}+\frac{1}{\sqrt{k-1}+\sqrt{k}} \\ =\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}=\sqrt{k+1}-\sqrt{k-1} \\ \Rightarrow 2\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\mathrm{...}+\frac{1}{\sqrt{79}+\sqrt{80}}\right)>\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\mathrm{...}+\sqrt{81}-\sqrt{79} \\ \Rightarrow 2\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\mathrm{...}+\frac{1}{\sqrt{79}+\sqrt{80}}\right)>9-1=8 \end{gathered}$$

Vậy $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\mathrm{...}+\frac{1}{\sqrt{79}+\sqrt{80}}>4$ .