Proof: Clearly, the function $f(t)=\sqrt{t}$ is convex on $(0,+\infty)$, so by Jensen's inequality we have
$$\begin{array}{l} \quad \sqrt{3 x+4}+\sqrt{6 y+4}+\sqrt{12 z+4} \leqslant 3 \cdot \\ \sqrt{\frac{(3 x+4)+(6 y+4)+(12 z+4)}{3}} \\ \quad=3 \cdot \sqrt{\frac{3(x+2 y+4 z)+12}{3}} \\ \quad=12 . \end{array}$$