ta có: \(\sqrt{27}+\sqrt{6}+1=3\sqrt{3}+\sqrt{6}+1\)(1))
\(\sqrt{48}=4\sqrt{3}=3\sqrt{3}+\sqrt{3}\)(2)
ta lại có: \(\sqrt{6}>\sqrt{3}\Rightarrow\sqrt{6}+1>\sqrt{3}\) (3)
từ (1)(2)và(3)\(\Rightarrow3\sqrt{3}+\sqrt{6}+1>3\sqrt{3}+\sqrt{3}\)
\(\Leftrightarrow\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)