Max of $(a-x^2)(b-y^2)(c-z^2)$ when $x+y+z=a+b+c=1$, $x,y,z,a,b,c \geq 0$

Question

What's the maximum value of

$$(a-x^2)(b-y^2)(c-z^2)$$

given $x+y+z=a+b+c=1$, $x,y,z,a,b,c \geq 0$

The tricky part, as you could see in one of the attempted answer, is how to handle the case when two of the 3 factors are negative

Answer 1

Partial solution

It is enough to consider the case where the product is positive.

1) If there are two negatives and one positive amongst $a-x^2, b-y^2, c-z^2$: I'm working on this.

2) If all $a-x^2, b-y^2, c-z^2 \geq 0$, then by applying AM-GM, we have

$$(a-x^2)(b-y^2)(c-z^2) \leq \Big( \frac{a+b+c-(x^2+y^2+z^2)}{3} \Big)^3 = \Big( \frac{1 -(x^2+y^2+z^2)}{3} \Big)^3$$

And $x^2+y^2+z^2 \geq \frac{(x+y+z)^2}{3} = \frac{1}{3}$. Thus, $$(a-x^2)(b-y^2)(c-z^2) \leq \frac{8}{729}$$

Equality happens when: $a-x^2 = b-y^2 = c-z^2 \geq 0$ and $x=y=z$, which means $x=y=z=1/3$ and $a=b=c=1/3$.