Cho A = 1/31 + 1/32 + 1/33 +...+ 1/60. Chứng minh rằng

* Hướng dẫn giải

Ta có

$$\begin{gathered} A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\mathrm{...}+\frac{1}{60} \\ =\left(\frac{1}{31}+\frac{1}{32}+\mathrm{....}+\frac{1}{44}+\frac{1}{45}\right)+\left(\frac{1}{46}+\frac{1}{47}+\mathrm{....}+\frac{1}{59}+\frac{1}{60}\right) \end{gathered}$$

Mà $\frac{1}{31}+\frac{1}{32}+\mathrm{....}+\frac{1}{44}+\frac{1}{45}>\frac{1}{45}+\frac{1}{45}+\mathrm{....}+\frac{1}{45}+\frac{1}{45}=\frac{15}{45}=\frac{1}{3}$

$\frac{1}{46}+\frac{1}{47}+\mathrm{....}+\frac{1}{59}+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\mathrm{....}+\frac{1}{60}+\frac{1}{60}=\frac{15}{60}=\frac{1}{4}$

Nên $A=\left(\frac{1}{31}+\frac{1}{32}+\mathrm{....}+\frac{1}{44}+\frac{1}{45}\right)+\left(\frac{1}{46}+\frac{1}{47}+\mathrm{....}+\frac{1}{59}+\frac{1}{60}\right)>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$

Vậy $A>\frac{7}{12}$