Đặt: f(x;y;z)=x^2+y^2+z^2+xyzTa 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=1Áp 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^2Suy 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}
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}$