a) \(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{60}\right)+...+\frac{1}{70}\)

Nhận xét:

\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\ge\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\ge\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)

\(\frac{1}{31}+...+\frac{1}{60}\ge\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{30}{60}=\frac{1}{2}\)

\(A\ge\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}...+\frac{1}{70}\ge\frac{1}{2}+\frac{1}{3}+\frac{1}{2}=\frac{4}{3}\)

Sorry ,tất cả dấu lớn hơn hoặc bằng đổi thành dấu > nhé 

còn câu b

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Đặt \(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)

\(=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{60}\right)+...+\frac{1}{70}\)

Nhận xét:

\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{30}{60}=\frac{1}{2}\)

\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}+...+\frac{1}{70}>\frac{1}{2}+\frac{1}{3}+\frac{1}{2}=\frac{4}{3}\)

\(\Rightarrow A>\frac{4}{3}\)

Vậy \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}>\frac{4}{3}\) (Đpcm)

\(A=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+....+\dfrac{1}{70}\\ =\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\right)+\left(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{30}\right)+\left(\dfrac{1}{30}+\dfrac{1}{31}+....+\dfrac{1}{60}\right)+....+\dfrac{1}{70}\\ \)

\(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}=\dfrac{1}{2}\)

\(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+....+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)

\(\dfrac{1}{30}+\dfrac{1}{31}+....+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{30}{60}=\dfrac{1}{2}\)

\(\Rightarrow A>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{61}+...+\dfrac{1}{70}>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{4}{3}\)

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