Example 1 If $x, y, z$ are positive real numbers, and satisfy $xyz = 1$, then the minimum value of the algebraic expression $(x+1)(y+1)(z+1)$ is ( . ).
(A) 64
(B) 8
(C) $8 \sqrt{2}$
(D) $\sqrt{2}$
(2002, Huanggang City, Hubei Province, Junior High School Mathematics Competition)
Solution: Let $M=(x+1)(y+1)(z+1)$,
then
$$
\begin{aligned}
M & =\frac{M}{1}=\frac{(x+1)(y+1)(z+1)}{x y z} \\
& =\frac{x+1}{x} \cdot \frac{y+1}{y} \cdot \frac{z+1}{z} \\
& =\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right) .
\end{aligned}
$$

From (1) and (2) we get
$$
\begin{aligned}
M^{2}= & \left(1+\frac{1}{x}\right)(x+1)\left(1+\frac{1}{y}\right)(y+1) . \\
& \left(1+\frac{1}{z}\right)(z+1) \\
= & \left(x+\frac{1}{x}+2\right)\left(y+\frac{1}{y}+2\right)\left(z+\frac{1}{z}+2\right) .
\end{aligned}
$$

But $x+\frac{1}{x}+2=\left(x-2+\frac{1}{x}+4\right)$
$$
=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+4 \text {. }
$$

Obviously, the minimum value of $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}$ is 0.
$$
\text { Therefore, }\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2}+4 \geqslant 4 \text {, }
$$

when $\sqrt{x}=\frac{1}{\sqrt{x}}$, the equality holds.
So, $x=1$.
Thus, $M^{2} \geqslant 4 \times 4 \times 4=64$, the minimum value is 64.
At this point, $x=y=z=1, M$ achieves the minimum value of 8.