Áp dụng \(\sqrt{\frac{a+b}{2}}>\frac{\sqrt{a}+\sqrt{b}}{2}\) được \(\sqrt{\frac{2007+2005}{2}}>\frac{\sqrt{2005}+\sqrt{2007}}{2}\Rightarrow2\sqrt{2006}>\sqrt{2005}+\sqrt{2007}\)
\(A=\sqrt{2005}+\sqrt{2007}\Rightarrow A^2=\left(\sqrt{2005}+\sqrt{2007}\right)^2=2005+2007+2\sqrt{2005\cdot2007}=4012+2\sqrt{\left(2006-1\right)\left(2006+1\right)}=4012+2\sqrt{2006^2-1}\)
\(B=2\sqrt{2006}\Rightarrow B^2=\left(2\sqrt{2006}\right)^2=4\cdot2006=2\cdot2006+2\cdot2006=4012+2\sqrt{2006^2}\)
Ta thấy \(4012=4012\) và \(\sqrt{2006^2-1}< \sqrt{2006^2}\)
nên \(A^2< B^2\)\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
Có \(\sqrt{2005}+\sqrt{2007}=2005+2007+2\sqrt{2005\cdot2007}\)
\(=2005+2007+2\sqrt{\left(2006-1\right)\left(2006+1\right)}\)
\(=4012+2\sqrt{2006^2-1}\)
\(2\sqrt{2006}=2006+2006+2\cdot2006\)
\(=4012+2\sqrt{2006^2}\)
Mà \(4012+2\sqrt{2006^2-1}< 4012+2\sqrt{2006^2}\)
\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)