Lời giải:
\((\sqrt{1993}+\sqrt{1995})^2=1993+1995+2.\sqrt{1993.1995}=3988+2\sqrt{(1994-1)(1994+1)}\)
\(=3988+2\sqrt{1994^2-1}< 3988+2\sqrt{1994^2}=3988+2.1994=7976\)
\(\Rightarrow \sqrt{1993}+\sqrt{1995}< \sqrt{7976}\) hay $\sqrt{1993}+\sqrt{1995}< 2\sqrt{1994}$