Maximum of the function $F(x,y,z):=\frac{x(1-x)y(1-y)z(1-z)}{1-(1-xy)z}$ I am stuck with finding the maximum of the function
\begin{align}
F(x,y,z):=\frac{x(1-x)y(1-y)z(1-z)}{1-(1-xy)z}
\end{align}
on the compact interval $[0,1]\times[0,1]\times[0,1].$
I think the Lagrange multiplier may work, but I have no idea to set a constraint function. Is there any hint I can follow or another method ? Any suggestions I will be grateful.
 A: Taking partial derivatives with respect to $x,y,z$ we have
$$\begin{cases}
\dfrac{\partial F}{\partial x} = -\dfrac{(y - 1) y (z - 1) z (x^2 y z - 2 x (z - 1) + z - 1)}{(z (x y - 1) + 1)^2}\\
\dfrac{\partial F}{\partial y} =-\dfrac{(x - 1) x (z - 1) z (x y^2 z - 2 y z + 2 y + z - 1)}{(x y z - z + 1)^2}\\
\dfrac{\partial F}{\partial z} =-\dfrac{(x - 1) x (y - 1) y (x y z^2 - z^2 + 2 z - 1)}{(x y z - z + 1)^2}
\end{cases}$$
Equating them to $0$ and dropping unnecessary factors which clearly don't lead to a maximum, we get
$$\begin{cases}
x^2 y z - 2 x (z - 1) + z - 1=0\\
x y^2 z - 2 y z + 2 y + z - 1=0\\
x y z^2 - z^2 + 2 z - 1 =0
\end{cases}$$
Then the desired solution is, according to WA
$$\begin{cases}
x = \sqrt{2} - 1\\ 
y = \sqrt{2} - 1\\
z = \frac{1}{\sqrt{2}}
\end{cases}$$
A: Take the logarithm, then just differentiate it.
EDIT
$$\frac{\partial}{\partial x}\ln F=\frac1x+\frac1{x-1}-\frac{yz}{1-z+xyz}=0 \\
\frac1y+\frac1{y-1}-\frac{xz}{1-z+xyz}=0 \\
\frac1z+\frac1{z-1}-\frac{xy-1}{1-z+xyz}=0$$
Multiply the first by $x$, the second by $y$ to get $x/(x-1)=y/(y-1)$.  Then turn the first equation into a linear equation in $z$.
A: Since $F(x,y,z)\ge0$ is continuous on a compact maximum and minimum exist. On the boundary it easy to check that $F=0$ therefore we can search the maximum by $\nabla F=0$ which leads to the condition
$$z=\frac1{1-\sqrt{xy}}=\frac{1-2x}{x^2y-2x+1} =\frac{1-2y}{y^2x-2y+1}$$
