Y Ribi Nkok Ngok Lê Nguyễn Ngọc Nhi Lê Anh Duy Nguyễn Thị Diễm Quỳnh trần thị diệu linh kudo shinichi Nguyen Giang Thủy Tiên Nguyễn Việt Lâm

Đặt \(C=\frac{1}{21}+\frac{1}{22}+....+\frac{1}{60}=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)

Ta có: \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};....\frac{1}{39}>\frac{1}{40}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+....+\frac{1}{39}+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{40}.20=\frac{1}{2}\) 

\(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...\frac{1}{59}>\frac{1}{60}\)

 \(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{1}{60}.20=\frac{1}{3}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}>\frac{11}{15}\)

Vậy \(C>\frac{11}{15}\) (1)

Lại có: \(\frac{1}{21}< \frac{1}{20};\frac{1}{22}< \frac{1}{20};...\frac{1}{40}< \frac{1}{20}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}+....+\frac{1}{20}=\frac{1}{20}.20=1\)

\(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...\frac{1}{60}< \frac{1}{40}\)

\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{40}.20=\frac{1}{2}\)

\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< \frac{1}{2}+1=\frac{3}{2}\)

Vậy \(C< \frac{3}{2}\) (2)

Từ (1) và (2) suy ra \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< \frac{3}{2}\)

Đề yêu cầu chứng tỏ gì hả bạn?

tách bất đẳng thức trên ta có \(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}\)gọi biều thức này là A

ta có \(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}\)

\(A=\left(\frac{20}{20.21}+\frac{21}{21.22}+\frac{22}{22.23}+...+\frac{39}{39.40}\right)+\left(\frac{40}{40.41}+\frac{41}{41.42}+...+\frac{59}{59.60}\right)\)

\(\Rightarrow A>20.\left(\frac{20}{20.21}+\frac{21}{21.22}+\frac{22}{22.23}+...+\frac{39}{39.40}\right)+40.\left(\frac{40}{40.41}+\frac{41}{41.42}+...+\frac{59}{59.60}\right)\)nhân vế trái vs 20 vế phải 40

\(\Rightarrow A>20.\left(\frac{1}{20}-\frac{1}{40}\right)+40.\left(\frac{1}{40}-\frac{1}{60}\right)\)

\(\Rightarrow A>\frac{5}{6}>\frac{11}{5}\left(1\right)\)

ta có \(A< 40.\left(\frac{20}{20.21}+\frac{21}{21.22}+\frac{22}{22.23}+...+\frac{39}{39.40}\right)+60.\left(\frac{40}{40.41}+\frac{41}{41.42}+...+\frac{59}{59.60}\right)\)

\(\Rightarrow A< 40.\left(\frac{1}{20}-\frac{1}{40}\right)+60.\left(\frac{1}{40}-\frac{1}{60}\right)\)

\(\Rightarrow A< \frac{3}{2}\left(2\right)\)

từ (1) và (2)

\(\Rightarrow\frac{11}{15}< A< \frac{3}{2}\)

\(\Rightarrow\frac{11}{15}< \text{​​}\text{​​}\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+..+\frac{1}{60}< \frac{3}{2}\)(ĐPCM)

Đáp án là mình chứng minh được.

Đặt  \(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}\)

=> \(A=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)

Đặt A < (1/40+.....+1/40)+(1/60+1/60+...+1/60)

=>A<1/2+1/3=5/6<3/2

lớn hơn 11/15 cũng tương tự thôi bạn tự làm sẽ thú vị hơn đấy

k minh nha

Thank you

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