\(\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}=1-\frac{1}{2004}+1-\frac{1}{2005}+1+\frac{2}{2003}\)
\(=3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)\)
Do \(\frac{1}{2003}>\frac{1}{2004}>\frac{1}{2005}.\) nên \(\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>0\)
Vì vậy \(3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>3\) (đpcm)
\(A=\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}\)
\(=(1-\frac{1}{2004})+(1-\frac{1}{2005})+(1+\frac{2}{2003})\)
\(=3+(\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005})\)
Do\(\frac{1}{2003}\)>\(\frac{1}{2004}\)>\(\frac{1}{2005}\)
\(\Rightarrow\frac{1}{2003}+\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\)>\(0\)
\(\Rightarrow3+(\frac{1}{2003}-\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2005})\)>\(3\)
\(\Rightarrow A\)>\(3\)