Đặt: $f(x;y;z)=x^2+y^2+z^2+4xyz$
Ta có: $f(x;y;z+t)\ge f(x;y;z),\forall t\ge0$, nên $\min A$ đạt được tại $x+y+z=1$.
Vì vậy, ta chỉ cần xét: $x+y+z=1$
Áp dụng BĐT AM-GM ta có:
$(x+y-z)(x-y+z)\le\dfrac{(x+y-z+x-y+z)^2}{4}=x^2$
$(x+y-z)(-x+y+z)\le\dfrac{(x+y-z-x+y+z)^2}{4}=y^2$
$(x-y+z)(-x+y+z)\le\dfrac{(x-y+z-x+y+z)^2}{4}=z^2$
Suy ra: $\left((x+y-z)(x-y+z)(-x+y+z)\right)^2\le(xyz)^2$
$\Rightarrow (x+y-z)(x-y+z)(-x+y+z)\le|(x+y-z)(x-y+z)(-x+y+z)|\le xyz$
$\Rightarrow (1-2z)(1-2y)(1-2x)\le xyz$
$\Leftrightarrow 1-2(x+y+z)+4(xy+yz+zx)-8xyz\le xyz$
$\Leftrightarrow xy+yz+zx\le\dfrac{9}{4}xyz+\dfrac{1}{4}$
Ta có:
$A=x^2+y^2+z^2+4xyz$
$=(x+y+z)^2-2(xy+yz+zx)+4xyz$
$\ge 1-2\left(\dfrac{9}{4}xyz+\dfrac{1}{4}\right)+4xyz$
$=\dfrac{1}{2}-\dfrac{1}{2}xyz$
$=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{(x+y+z)^3}{27}=\dfrac{13}{27}$
$\min A=\dfrac{13}{27} \Leftrightarrow x=y=z=\dfrac{1}{3}$