S = \(\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{130}\right)+\left(\frac{1}{111}+...+\frac{1}{120}\right)+\left(\frac{1}{121}+...+\frac{1}{130}\right)\)
\(\Rightarrow\frac{1}{110.10}+\frac{1}{120.10}+\frac{1}{130.10}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}>\frac{1}{12}+\frac{2}{12}=\frac{1}{4}\)
= \(\frac{1}{11}+\frac{1}{13}>\frac{2}{12}\)
\(\Rightarrow S>\frac{1}{4}\left(1\right)\)
\(S=\left(\frac{1}{101}+\frac{1}{130}\right)+\left(\frac{1}{102}+\frac{1}{129}\right)+...+\left(\frac{1}{115}+\frac{1}{116}\right)\)( có 15 cặp )
\(=\frac{231}{101.130}+\frac{231}{102.129}+...+\frac{231}{115.116}=231\)
\(\left(\frac{1}{101.130}+\frac{1}{102.129}+...+\frac{1}{115.116}\right)\)
Ta nhận xét tích 101.130 có giá trị nhỏ nhất :
xét : 102.129 = (101+1).(130-1) = 101.129 = 101.130 - 101 + 130 - 1 = 101.130 + 28 > 101.130
Tương tự các cặp cộng lại , ta có : \(\frac{1}{101.130}+\frac{1}{129.102}+...+\frac{1}{115.116}< \frac{1}{101.130.15}\)
\(\Rightarrow S=\frac{231.1}{101.130.15}=\frac{693}{2626}< \frac{91}{330}\)
Từ (1)(2) \(\Rightarrow\)ĐPCM
