Ta có : \(\frac{7}{12}=\frac{4}{12}+\frac{3}{12}=\frac{1}{3}+\frac{1}{4}\)
Ta chia tổng S thành 2 tổng nhỏ hơn như sau :
\(S=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{79}+\frac{1}{80}=\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)
+) Vì \(\frac{1}{41}>\frac{1}{42}>\frac{1}{43}>...>\frac{1}{60}\) => \(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(\Rightarrow\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)>\frac{1}{60}\times20\)
\(\Rightarrow\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)>\frac{1}{3}\)
+) Vì \(\frac{1}{61}>\frac{1}{62}>\frac{1}{63}>...>\frac{1}{80}\Rightarrow\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\)
\(\Rightarrow\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)>\frac{1}{80}\times20\)
\(\Rightarrow\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)>\frac{1}{4}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{3}+\frac{1}{4}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{7}{12}\)
Vậy \(S>\frac{7}{12}\) ( đpcm )
