29. Given that the real numbers $x, y$ and $z$ satisfies the condition $x+y+z=3$, find the maximum value of $f(x y z)=\sqrt{2 x+13}+\sqrt[3]{3 y+5}+\sqrt[4]{8 z+12}$.
29. Answer: 8
Using the AM-GM inequality, we have
$$
\begin{aligned}
f(x y z) & =\sqrt{2 x+13}+\sqrt[3]{3 y+5}+\sqrt[4]{8 z+12} \\
& =\sqrt{\frac{2 x+13}{4}} \cdot \sqrt{4}+\sqrt[3]{\frac{3 y+5}{4}} \cdot \sqrt[3]{2} \cdot \sqrt[3]{2}+\sqrt[4]{\frac{8 z+12}{8}} \cdot \sqrt[4]{2} \cdot \sqrt[4]{2} \cdot \sqrt[4]{2} \\
& \leq \frac{\frac{2 x+13}{4}+4}{2}+\frac{\frac{3 y+5}{4}+2+2}{3}+\frac{\frac{8 x+12}{8}+2+2+2}{4} \\
& =\frac{1}{4}(x+y+z)+\frac{29}{4} \\
& =8
\end{aligned}
$$

The equality is achieved at $x=\frac{3}{2}, y=1$ and $z=\frac{1}{2}$.