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$x,y,z\geq 0; x+y+z\geq 1$ . Tìm min

$A=x^{2}+y^{2}+z^{2}+4xyz$

khangnguyenthanh · Answer 1 · 10/22/2013 12:37:34 PM

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:

$A=x^2+y^2+z^2+4xyz$

$=(x+y+z)^2-2(xy+yz+zx)+4xyz$

$\ge 1-2\left(\dfrac{9}{4}xyz+\dfrac{1}{4}\right)+4xyz$

$=\dfrac{1}{2}-\dfrac{1}{2}xyz$

$=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{(x+y+z)^3}{27}=\dfrac{13}{27}$

$\min A=\dfrac{13}{27} \Leftrightarrow x=y=z=\dfrac{1}{3}$

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