2)
\(\left(\sqrt{2003}+\sqrt{2005}\right)^2=2003+2005+2\sqrt{2003\times2005}\)\(=4008+2\sqrt{\left(2004-1\right)\left(2004+1\right)}=4008+2\sqrt{2004^2-1}\)
\(\left(\sqrt{2004}+\sqrt{2004}\right)^2=2004+2004+2\sqrt{2004\times2004}\)\(=4008+2\sqrt{2004^2}\)
Ta có \(2004^2>2004^2-1\Rightarrow\sqrt{2004^2}>\sqrt{2004^2-1}\Rightarrow4008+2\sqrt{2004^2}>4008+2\sqrt{2004^2-1}\)
Vậy \(2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)