Extend 1 Let non-negative real numbers $a, b, c, x, y, z$ satisfy $a+b+c=x+y+z=1$.
Find the minimum value of $\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right)$.
【Analysis】When $a=y=z=0, x=1, b=c=\frac{1}{2}$, the original expression is $=-\frac{1}{4}$.
Below is the proof:
$$
\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right) \geqslant-\frac{1}{4} \text {. }
$$
(1) If $a-x^{2} 、 b-y^{2} 、 c-z^{2}$ are all negative, then
$$
\begin{array}{l}
a-\frac{1}{4} \text {. }
$$
(3) If one of $a-x^{2} 、 b-y^{2} 、 c-z^{2}$ is negative and the other two are positive, without loss of generality, assume $a-x^{2}<0$. Then
$$
\begin{array}{l}
-\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right) \\
=\left(x^{2}-a\right)\left(b-y^{2}\right)\left(c-z^{2}\right) \\
\leqslant x^{2} b c \leqslant 1 \cdot\left(\frac{b+c}{2}\right)^{2} \leqslant \frac{1}{4},
\end{array}
$$

Equality holds if and only if $a=0, b=c=\frac{1}{2}, x=1, y=z=0$.
Therefore, $\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right) \geqslant-\frac{1}{4}$.
In summary, the minimum value sought is $-\frac{1}{4}$.