Bugi wrote: Let $ x,y,z$ be positive real numbers such that $ x + 2y + 3z = \frac {11}{12}$. Prove the inequality $ 6(3xy + 4xz + 2yz) + 6x + 3y + 4z + 72xyz\le \frac {107}{18}$. Let $ P = 6(3xy + 4xz + 2yz) + 6x + 3y + 4z + 72xyz = 12(x+\frac{1}{6})(2y+\frac{2}{3})(3z+\frac{3}{4}) - 1$ Then by AM-GM: $ P \le 12(\frac{x+\frac{1}{6} + 2y + \frac{2}{3} + 3z + \frac{3}{4}}{3})^3 - 1$ And we know: $ \frac{x+\frac{1}{6} + 2y + \frac{2}{3} + 3z + \frac{3}{4}}{3} = \frac{5}{6}$ So $ P \le 12(\frac{5}{6})^3 = \frac{125}{8} - 1 = \frac{107}{18}$ Equality with $ x = \frac{2}{3}, y = \frac{1}{12}, z = \frac{1}{36}$