Có : 2 > \(\sqrt{3}\) ; 3 > \(\sqrt{4}\) ; ..... ; 1999 > \(\sqrt{2000}\)
=> VT = \(\sqrt{2\sqrt{3\sqrt{4......\sqrt{1999\sqrt{2000}}}}}\)< \(\sqrt{2\sqrt{3\sqrt{4......\sqrt{1999.1999}}}}\)
= \(\sqrt{2\sqrt{3\sqrt{4.....\sqrt{1999}}}}\) < ........ < \(\sqrt{2\sqrt{3}}\) < \(\sqrt{2.2}\) = 2
=> ĐPCM
Ta có: \(n=\sqrt{n^2}=\sqrt{1+n^2-1}=\sqrt{1+n-1.n+1}\)
Áp dụng công thức trên với \(n=4,5,6\)ta có:
\(4=\sqrt{1+3.5}=\sqrt{1+3\sqrt{1+4\sqrt{1+5.7}}}=\sqrt{1+3\sqrt{1+\sqrt{4\sqrt{1+...n-1\sqrt{n+1^2}}}}}\)
\(>\sqrt{3\sqrt{4\sqrt{...2000}}}\)
Do đó: \(\sqrt{2+\sqrt{3\sqrt{4\sqrt{...2000}}}}< \sqrt{2+2}=2\)
