Saudi Arabia Mathematical Competitions 2012

For any positive real numbers $x,y,z$ we have $$\begin{array}{rl}\frac{x+1}{y}+\frac{2y+1}{z}+\frac{3z+1}{x}& =\frac{x}{y}+\frac{2y}{z}+\frac{3z}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\\ & \ge 3\sqrt{6}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>3\sqrt{6}.\end{array}$$ Hence $a\ge 3\sqrt{6}$.

Consider the triples $(x,y,z)$ satisfying the equality case in the previous AM-GM inequality. We have $$\frac{x}{y}=\frac{2y}{z}=\frac{3z}{x}=\sqrt{6}.$$ It follows that $x=\frac{3}{\sqrt{6}}z$, $y=\frac{\sqrt{6}}{2}z$. So all the desired triples are given by $\left(\frac{3}{\sqrt{6}}z,\frac{\sqrt{6}}{2}z,z\right)$, and we obtain $$3\sqrt{6}+\left(\frac{\sqrt{6}}{3}+\frac{2}{\sqrt{6}}+1\right)\cdot \frac{1}{z}>a,$$ For $z\to +\infty$, we deduce that $3\sqrt{6}\ge a$.

Thus, the answer is $a=3\sqrt{6}$.