Question

Chứng minh rằng: 1\41+1\42+...+1\79+1\80>7\12

 

Answer 1

CMR 1/41 + 1/42 + 1/43 + ... + 1/79 + 1/80 > 7/12 

Ta có: 
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80 

1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80) 

Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60 
\(\Rightarrow\) (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60 

và 1/61> 1/62> ... >1/79> 1/80 
\(\Rightarrow\) (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80 

Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12 

\(\Rightarrow\) 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12 

\(\RightarrowĐPCM\)

Answer 2

Đặt S = \(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{79}+\frac{1}{80}=\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)

Ta có: \(\frac{1}{41}>\frac{1}{60}\)

\(\frac{1}{42}>\frac{1}{60}\)

..............

\(\frac{1}{59}>\frac{1}{60}\)

\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{59}+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)(1)

Lại có: \(\frac{1}{61}>\frac{1}{80}\)

\(\frac{1}{62}>\frac{1}{80}\)

............

\(\frac{1}{79}>\frac{1}{80}\)

\(\Rightarrow\frac{1}{61}+\frac{1}{62}+...+\frac{1}{79}+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}+\frac{1}{80}=\frac{20}{80}=\frac{1}{4}\)(2)

Lấy (1) + (2) ta được:

\(S>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

Vậy S > 7/12 (ĐPCM)