The answer is $\boxed{\frac{1}{16}}$, which is achieved at $(a, b, c, x, y, z) = (1, 0, 0, 0, \frac12, \frac12).$ Now let's show this is optimal. Note that it suffices to consider the cases when an odd number of $a-x^2, b-y^2, c-z^2$ are positive and the rest are negative. We consider two relevant cases, WLOG. Case 1. $a-x^2, b-y^2, c-z^2 >0$ By AM-GM, we have that $(a-x^2)(b-x^2)(c-z^2) \le (\frac{a + b + c - x^2 - y^2 - z^2}{3})^3 = (\frac{1 - x^2 - y^2 - z^2}{3})^3 \le (\frac{\frac23}{3})^3 = \frac{8}{729} \le \frac{1}{16}$, as claimed. Case 2. $a-x^2 > 0, b-y^2, c-z^2 < 0$ In this case, it's clearly optimal to have $a = 1, b = c = 0$, and so we wish to optimize $(1-x^2)y^2z^2$ subject to $x+y+z = 1.$ By AM-GM, we must have $y = z$, and so we seek to optimize $(1-x^2)(\frac{1-x}{2})^2$ where $x \in (0, 1).$ Setting the derivative of this w.r.t. $x$ to $0$ yields that $(1-x^2) (2x-2) + (-2x)(1-x)^2 = 0.$ This yields either $x = 1$ or $2x+1 = 0$, and so this is maximized at either $x = 1$ or $x = 0$ on $[0, 1]$. Clearly this is optimal at $x = 0$, and so we have that $(a-x^2)(b-x^2)(c-z^2) \le \frac{1}{16}$ in this case too. $\square$

This is an old problem which was posted my sqing quite back

minecraftfaq wrote: For nonnegative real numbers $a,b,c,x,y,z$, if$a+b+c=x+y+z=1$, find the maximum value of $(a-x^2)(b-y^2)(c-z^2)$. https://artofproblemsolving.com/community/c6h1892042p12913610