Đặt $A = \dfrac{15}{\sqrt[]{6}+1} +\dfrac{4}{\sqrt[]{6}-2}$
$ = \dfrac{15.(\sqrt[]{6}-1)}{6-1}+\dfrac{4.(\sqrt[]{6}+2)}{6-4}$
$ = 3.(\sqrt[]{6}-1) +2.(\sqrt[]{6}+2)$
$ = 5\sqrt[]{6} + 1$
Có $B = \dfrac{12}{3-\sqrt[]{6}}+\sqrt[]{6}$
$ = \dfrac{12.(3+\sqrt[]{6})}{9-6}+\sqrt[]{6}$
$ = 4.(3+\sqrt[]{6}) + \sqrt[]{6}$
$ = 5\sqrt[]{6}+12$
Dễ thấy $A<B$
Nên $\dfrac{15}{\sqrt[]{6}+1} +\dfrac{4}{\sqrt[]{6}-2} < \dfrac{12}{3-\sqrt[]{6}}+\sqrt[]{6}$