Ta có: \(\frac{2017}{2018}+\frac{2018}{2017}\)
\(=\frac{2017}{2018}+\frac{1}{2017}+1\)
Lại có: \(\frac{1}{2017}>\frac{1}{2018}\)
\(\Rightarrow\frac{2017}{2018}+\frac{1}{2017}+1>\frac{2017}{2018}+\frac{1}{2018}+1\)\(=1+1=2\)
\(\Rightarrow\frac{2017}{2018}+\frac{2018}{2017}>2\)
Vậy: \(\frac{2017}{2018}+\frac{2018}{2017}>2\)