<s>[INST] $xy + yz + zx + 2xyz = 1$ 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.
 [/INST] $x(y+z)+yz(1+2x)=1,yz\le (\dfrac{y+z}{2})^2 ,p=y+z \implies xp+(1+2x)\dfrac{p^2}{4} \ge 1 \\ \implies x \ge  \dfrac{1-\frac{p^2}{4}}{p+\frac{p^2}{2}}=\dfrac{1}{p}-\dfrac{1}{2}$
$4x+y+z \ge \dfrac{4}{p}-2+p \ge 2\sqrt{\dfrac{4}{p}*p}-2=2$
</s>