81. Let positive real numbers $x, y, z$ satisfy the condition $2 x y z=3 x^{2}+4 y^{2}+5 z^{2}$, find the minimum value of the expression $P=3 x+2 y+z$.


Solution: Let $a=3 x, b=2 y, c=z$, then we have
$$a+b+c=3 x+2 y+z, \quad a^{2}+3 b^{2}+15 c^{2}=a b c$$

According to the weighted AM-GM inequality, we have
$$\begin{array}{l}
a+b+c \geq(2 a)^{\frac{1}{2}}(3 b)^{\frac{1}{3}}(6 c)^{\frac{1}{6}} \\
a^{2}+3 b^{2}+15 c^{2} \geq\left(4 a^{2}\right)^{\frac{1}{4}}\left(9 b^{2}\right)^{\frac{3}{9}}\left(36 c^{2}\right)^{\frac{15}{36}}=\left(4 a^{2}\right)^{\frac{1}{4}}\left(9 b^{2}\right)^{\frac{1}{3}}\left(36 c^{2}\right)^{\frac{5}{12}}
\end{array}$$

Multiplying the above inequalities, we have
$$(a+b+c)\left(a^{2}+3 b^{2}+15 c^{2}\right) \geq 36 a b c \Rightarrow a+b+c \geq 36$$

Therefore, the minimum value of $3 x+2 y+z$ is 36. This is achieved when $x=y=z=6$.