Minimize (2,1,0) with constraint (2,1,1)

let $x + y = a , y + z = b , x + z = c$ and $s = \large{\frac{a + b + c}{2}}$

then the given constraint becomes $(s-a)(s-b)(s-c)s = 1$

$a,b,c$ can be treated as the sides of triangle.

let $A,B,C$ be the angles opposite to sides $a,b,c$ .

${\frac{1}{2}} ab \sin C = 1$

${\frac{1}{2}} bc \sin A = 1$

${\frac{1}{2}} ac \sin B = 1$

multiplying them all and we get :

${\frac{1}{8}} (abc)^2 \sin A \sin B \sin C = 1$

by A.M - G.M and Jensen's inequality

$\sqrt[3]{\sin A \sin B \sin C} \leq \frac{\sin A + \sin B + \sin C}{3} \leq \sin \frac{A + B + C}{3}$

So, $\sin A \sin B \sin C \leq \large\frac{3\sqrt{3}}{8}$

Thus , $(abc)^2 \geq \frac{64}{3\sqrt{3}}$ or $abc \geq \boxed{3.50953}$

also, equality occcurs at $x = y = z = \frac{1}{\sqrt[4]{3}}$

By AM-GM inequality, $\frac{x+y+z}{3} \geq \sqrt[3]{xyz} \Rightarrow x+y+z \geq 3 \sqrt[3]{xyz},$ so $1 = xyz(x+y+z) \geq xyz \cdot 3 \sqrt[3]{xyz} \Rightarrow xyz \geq \left(\frac{1}{3}\right)^{3/4}$ .

Expanding $(x+y)(y+z)(z+x) = 2xyz + \displaystyle\sum_{\textrm{cyc}} x^2 y$ . Applying AM-GM on $\displaystyle\sum_{\textrm{cyc}} x^2 y$ , $\frac{\displaystyle\sum_{\textrm{cyc}} x^2 y}{6} \geq xyz \Rightarrow \sum_{\textrm{cyc}} x^2 y \geq 6xyz.$ Therefore, we have $(x+y)(y+z)(z+x) = 2xyz + \sum_{\textrm{cyc}} x^2 y \geq 8xyz.$ Finally, $(x+y)(y+z)(z+x) \geq 8xyz \geq 8 \left(\frac{1}{3}\right)^{3/4}$ where equality holds in the inequalities if and only if $x=y=z = \left(\frac{1}{3}\right)^{1/4}$ .

The minimum can be found by these inequalites: $(x+y)(y+z)(z+x)\geq\frac{8}{9}(x+y+z)(xy+yz+xz)$ , $(xy+yz+xz)^2\geq 3xyz(x+y+z)$ and $(x+y+z)^2\geq 3(xy+yz+xz)$

$\displaystyle xyz(x+y+z)=1 \implies xyz(3\sqrt[3]{xyz}) \le 1 \implies xyz \le \frac{1}{\sqrt[4]{27}}$

$\displaystyle (x+y)(y+z)(z+x)=(x+y+z)(xy+yz+xz)-xyz$

$\displaystyle (x+y+z)(xy+yz+xz) \ge 3(x+y+z)\sqrt[3]{x^2y^2z^2} = \frac{3(x+y+z)xyz}{\sqrt[3]{xyz}} = \frac{3}{\sqrt[3]{xyz}} \ge 3\sqrt[4]{3}$

$\displaystyle \implies (x+y)(y+z)(z+x) \ge 3\sqrt[4]{3}-\frac{1}{\sqrt[4]{27}}=\frac{8\sqrt[4]{3}}{3}$