Đặt A = \(\sqrt{ }\)2003 + \(\sqrt{ }\)2005 ; B = 2\(\sqrt{ }\)2004
A² = 2003 + 2005 + 2\(\sqrt{ }\)(2003.2005)
= 4008 + 2\(\sqrt{ }\)[(2004-1)(2004+1)]
= 4008 + 2\(\sqrt{ }\)(2004² - 1) < 2.2004 + 2\(\sqrt{ }\)(2004²) = 4.2004 = B²
\(\Rightarrow\) A < B
Ta có: \(2\sqrt{2003.2005}=2\sqrt{2004^2-1}< 2\sqrt{2004^2}\)
\(\Rightarrow\) 2003 + \(2\sqrt{2003.2005}+2005\) < 2003 + 4008 + 2005
hay \(\left(\sqrt{2003}+\sqrt{2005}\right)^2< 8016\)
\(\Rightarrow\) \(\sqrt{2003}+\sqrt{2005}\) < 2 \(\sqrt{2004}\)