Is it just AM-GM?

We have: $\dfrac{x}{1+x}=1-\dfrac{y}{1+y}+1-\dfrac{z}{1+z}=\dfrac{1}{1+y}+\dfrac{1}{1+z}\ge2\sqrt{\dfrac{1}{(1+y)(1+z)}}$ .

Similarly, we get $\dfrac{y}{1+y}\ge2\sqrt{\dfrac{1}{(1+x)(1+z)}};\dfrac{z}{1+z}\ge2\sqrt{\dfrac{1}{(1+x)(1+y)}}$ .

Hence, $\dfrac{xyz}{(1+x)(1+y)(1+z)}\ge\dfrac{8}{(1+x)(1+y)(1+z)}$ or $xyz\ge8$ .

Equality occurs when $x=y=z=2$ .

expanding $\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}=2$ we get $xyz=x+y+z+2$ . Now we use AM-GM

$\begin{aligned} xyz&=x+y+z+2\\ &\ge 3\sqrt[3]{xyz}+2 \\ t^3&=3t+2\text{ where }t=\sqrt[3]{xyz} \\ 0&\le t^3-3t-2\\ 0 &\le (t+1)^2(t-2)\end{aligned}$

because $(t+1)^2$ is always positive for all real $x,y,z$ we must have $\sqrt[3]{xyz}=t\ge 2$ so $xyz\ge 8$ and equality occurs when $x=y=z=2$

First, we note that $\frac{x}{1+x} = 1 -\frac{1}{1+x}$ and so are the other equations. Then, we have

$1 - \frac{1}{1+x} + 1 -\frac{1}{1+y} + 1 - \frac{1}{1+z} = 2$

$\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} = 1$ .

By AM-HM inequality, we have

$\frac{1}{3} \quad \geq \quad \frac{3}{x+1+y+1+z+1}$

Simplify, we have $x + y + z \space \geq 6$ .

Equality occurs when $x=y=z=2$ .

Thus, $xyz= 2 \times 2 \times 2 = \boxed{8}$ .

For the minimum value, we have to make x=y=z. So we have 3(x)/(1+x) = 2.Then x=2. So the minimum value of xyz=8