Ta có:
8/9=1/9+1/9+1/9+1/9+1/9+1/9+1/9+1/9
Mà 1/9<1/2;1/9<1/3;...1/9<1/8;1/9=1/9
=>1/9+1/9+...+1/9<1/2+1/3+...+1/8+1/9
Vậy.......
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\)>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=\frac{1}{1}-\frac{1}{9}\)
\(=\frac{8}{9}\)
Nên > \(\frac{8}{9}\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>1-\frac{1}{9}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>\frac{8}{9}\)
\(\RightarrowĐPCM\)