$xy + yz + zx + 2xyz = 1$ implies $4x+y+z\geq 2$ Let $x,y,z>0$ satisfy $$xy + yz + zx + 2xyz = 1.$$ Prove that $4x+y+z\geq 2$.
The condition invites the factoring $(1+x)(1+y)(1+z)+xyz-2=x+y+z$, but having the factor $4$ in the desired inequality makes things more difficult.
 A: since 
$$xy+yz+xz+2xyz=1$$
Note this following indentity
$$\sum_{cyc}\dfrac{ab}{(b+c)(c+a)}+\dfrac{2abc}{(a+b)(b+c)(c+a)}=1$$
so we Let $$x=\dfrac{a}{b+c},y=\dfrac{b}{c+a},z=\dfrac{c}{a+b},a,b,c>0$$
$$\Longleftrightarrow \dfrac{4a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge 2\tag{1}$$
Use Cauchy-Schwarz inequality we have
$$\left(\dfrac{4a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\left(a(b+c)+b(c+a)+c(a+b)\right)\ge (2a+b+c)^2$$
then it suffices prove
$$(2a+b+c)^2\ge 4(ab+bc+ac)$$
$$4a^2+b^2+c^2-2bc\ge 0$$
$$4a^2+(b-c)^2\ge 0$$
It is clear.
Proof $(1)$ other methods
let $$b+c=x,c+a=y,a+b=z$$
then
$$a=\dfrac{y+z-x}{2},b=\dfrac{x+z-y}{2},c=\dfrac{x+y-z}{2}$$
then inequality $(1)$ can write
$$\Longleftrightarrow \dfrac{2(y+z-x)}{x}+\dfrac{x+z-y}{2y}+\dfrac{x+y-z}{2z}\ge 2$$
$$\Longleftrightarrow \left(\dfrac{2y}{x}+\dfrac{x}{2y}\right)+\left(\dfrac{2z}{x}+\dfrac{x}{2z}\right)+\left(\dfrac{z}{2y}+\dfrac{y}{2z}\right)-3\ge 2$$
Use AM-GM inequality By Done
A: $x(y+z)+yz(1+2x)=1,yz\le (\dfrac{y+z}{2})^2 ,p=y+z \implies xp+(1+2x)\dfrac{p^2}{4} \ge 1 \\ \implies x \ge  \dfrac{1-\frac{p^2}{4}}{p+\frac{p^2}{2}}=\dfrac{1}{p}-\dfrac{1}{2}$
$4x+y+z \ge \dfrac{4}{p}-2+p \ge 2\sqrt{\dfrac{4}{p}*p}-2=2$
