, 2012

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

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 $3\sqrt{6} \geq a$.

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