The 6th problem of the 3rd China Northern Hope Star Mathematics Summer Camp:
Given non-negative real numbers $a, b, c, x, y, z$ satisfying $a+b+c=x+y+z=1$.
Find the maximum value of $\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right)$.
【Analysis】When $a=b=z=0, c=1, x=y=\frac{1}{2}$, the original expression $=\frac{1}{16}$.
Below is the proof: $\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right) \leqslant \frac{1}{16}$.
(1) If $a-x^{2} 、 b-y^{2} 、 c-z^{2}$ are all non-negative, then according to the mean inequality we get
$$
\begin{array}{l}
\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right)<a b c \\
\leqslant\left(\frac{a+b+c}{3}\right)^{3}=\frac{1}{27}<\frac{1}{16} .
\end{array}
$$
(2) If one or three of $a-x^{2} 、 b-y^{2} 、 c-z^{2}$ are negative, then
$$
\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right) \leqslant 0<\frac{1}{16} \text {. }
$$
(3) If exactly two of $a-x^{2} 、 b-y^{2} 、 c-z^{2}$ are negative, without loss of generality, assume $a-x^{2}<0, b-y^{2}<0, c-z^{2} \geqslant 0$. Then
$$
\begin{array}{l}
\left|\left(a-x^{2}\right)\left(b-y^{2}\right)\right|=\left(x^{2}-a\right)\left(y^{2}-b\right) \\
\leqslant x^{2} y^{2} \leqslant\left(\frac{x+y}{2}\right)^{4} \leqslant \frac{1}{16},
\end{array}
$$

The equality in (1) holds if and only if $x=y=\frac{1}{2}, a=b=0$; $\square$
$$
\left|c-z^{2}\right| \leqslant c \leqslant 1,
$$

The equality in (1) holds if and only if $c=1, z=0$.
Therefore, $\left(a-x^{2}\right)\left(b-y^{2}\right)\left(c-z^{2}\right) \leqslant \frac{1}{16}$.
In summary, the maximum value sought is $\frac{1}{16}$.