A=\(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\) +\(\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)
Ta có : \(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...;\frac{1}{60}=\frac{1}{60}\) => \(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}>\frac{20}{60}=\frac{1}{3}\)
\(\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};...;\frac{1}{80}=\frac{1}{80}\) => \(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{20}{80}=\frac{1}{4}\)
=> A > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
Vậy a >\(\frac{7}{12}\)
\(\frac{7}{12}=\frac{3}{12}+\frac{4}{12}=\frac{1}{4}+\frac{1}{3}\)
ta có:\(A=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}=\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)
ta có:\(\frac{1}{41}>\frac{1}{42}>\frac{1}{43}>...>\frac{1}{60}\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{59}+\frac{1}{60}>\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\left(1\right)\)
\(\frac{1}{61}>\frac{1}{62}>\frac{1}{63}>...>\frac{1}{80}\Rightarrow\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{20}{80}=\frac{1}{4}\left(2\right)\)
từ (1) (2) suy ra \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}>\frac{7}{12}\left(đfcm\right)\)
