$x, y, z$ are real numbers which satisfy the following:
$$x + xy + xyz = 1$$
$$y + yz + xyz = 2$$
$$z + zx + xyz = 4$$Then find the maximum value of $xyz$.
I tried adding and subtracting a few equations and substituting iteratively into the others which didn't work. Then I tried letting $xyz = k$ and trying to get stuff in terms of $k$ but I couldn't proceed with that either.
Best Answer
I directly solved the three equations. From first equation $$ z = \frac{1-x}{xy}-1 = \frac{1}{xy}-\frac{1}{y}-1$$, substituting this in the second equation, $$ y + y\left(\frac{1-x}{xy}-1 \right) + (1-x-xy)=2 \implies xy = \frac{1}{x}-x-2. $$ This implies $xyz = 1-x-xy = 3-\frac{1}{x}$ and $z= \frac{1-x}{\frac{1}{x}-x-2}-1 = \frac{3x-1}{1-x^2-2x}$. Substituting this in the third equation results in a cubic equation, $$\left(\frac{3x-1}{1-x^2-2x}\right)+\left(\frac{3x^2-x}{1-x^2-2x}\right)+\left(3-\frac{1}{x}\right)=4 \\ \implies \frac{3x^2+2x-1}{1-x^2-2x}-\frac{1}{x}=1 \implies 3x^2+2x-1-\frac{1}{x}+x+2 = 1-x^2-2x\\ \implies \boxed{4x^3+5x^2-1=0}. $$ This has three solutions $x = -1,-\frac{1}{8}-\frac{\sqrt{17}}{8},-\frac{1}{8}+\frac{\sqrt{17}}{8}$. The maximum of $xyz = 3-\frac{1}{x}$ is attained when $x=-\frac{1}{8}-\frac{\sqrt{17}}{8}$ and it is $$ \boxed{\frac{1}{2}(5+\sqrt{17})} $$