Prove maximum value of $(z-xy)(x-yz)(y-zx)$ is $\frac{1}{64}$ given $x,y,z \in (0,1)$ Prove maximum value of $(z-xy)(x-yz)(y-zx)$ is $\frac{1}{64}$ given $x,y,z \in (0,1)$
I can make it $\frac{1}{64}$ by setting $x,y,z = \frac{1}{2}$, but I have no idea how to show that's the maximum.
 A: As it happens, just yesterday I showed on this forum, 
in the process of answering
another
stackexchange question, that if we fix the average
$a=\frac13(x+y+z)$ then
$$
\left( 1 - \frac{xy}{z} \right)
\left( 1 - \frac{yz}{x} \right)
\left( 1 - \frac{xz}{y} \right)
\leq (1-a)^3
$$
with equality iff $x=y=z=a$.
Also, by the inequality on arithmetic and geometric means
(which also figured in the proof of the $(1-a)^3$ bound),
we have $xyz \leq a^3$, again with equality iff $x=y=z=a$.
Multiplying these two inequalities yields
$$
(z - xy) (x - yz) (y - xz) \leq a^3 (1-a)^3 = \bigl( a(1-a) \bigr)^3,
$$
and one final application of the AM-GM inequality shows that
this is at most $(1/4)^3 = 1/64$ with equality iff a=1/2, QED.
A: Here is a cheap way. Denote the LHS by $P$. First note that all three factors must be positive to get a positive product. 
Now write $(x-yz)(y-xz)=xy(1+z^2)-(x^2+y^2)z\le xy(1+z^2)-2xyz=xy(1-z)^2$.
Multiply by $z-xy$ to get
$$
P\le xy(1-z)^2(z-xy)\,.
$$ 
Since everything is symmetric, we can just as well write
$$
P\le yz(1-x)^2(x-yz)
$$
and 
$$
P\le xz(1-y)^2(y-xz)\,.
$$ 
Multiplying these out, we get
$$
P^3\le [xyz(1-x)(1-y)(1-z)]^2P\,
$$ 
so $P\le xyz(1-x)(1-y)(1-z)$. However, $x(1-x)=\frac 14-(x-\frac 12)^2\le\frac 14$ and the same is true for the other two products.
A: Well you can think it in this way. Let's suppose it has a maximum (the problem ask to find it, so I think we can safely assume it has one). Now give the symmetry of the function is easy to see that if you change any of the 3 variables the function does not change. Geometrically it means you could change and of the 3 axis and the graph would not change. That means that the maximum must be at $x=y=z$. Otherwise you could exchange two axis and the graph would be changed (rotated) and the maximum would be shifted (think in 2 dimension to an oval and try to see what happens if you exchange $x$ with $y$), but the function will not change so this is not possible. Now given the fact that the $x,y,z$ are equal let's call them $x=y=z=s$, our equation will become
$$
(s-s^2)^3=\frac{1}{64}
$$
that means
$$
s-s^2=1/4
$$
and therefore given the conditions $s=1/2$.
I hope is understandable.
