\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\)
= \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2013}{2013}+\frac{1}{2013}+\frac{1}{2013}\)
= \(\left(\frac{2013}{2014}+\frac{1}{2013}\right)+\left(\frac{2014}{2015}+\frac{1}{2013}\right)+1\)
Ta có :\(\frac{2013}{2014}+\frac{1}{2013}>\frac{2013}{2014}+\frac{1}{2014}=1\)
\(\frac{2014}{2015}+\frac{1}{2013}>\frac{2014}{2015}+\frac{1}{2015}=1\)
=> A > 1 + 1 + 1 = 3
Vậy A > 3