Đặt: f(x;y;z)=x2+y2+z2+4xyz
Ta có: f(x;y;z+t)≥f(x;y;z),∀t≥0, nên min 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}