Cho x,y,z thoả mãn x+y+z=1. Tìm min
$M=\sqrt{x^{2}+xy+y^{2}}+\sqrt{y^{2}+yz+z^{2}}+\sqrt{z^{2}+zx+x^{2}}$
Cho x,y,z thoả mãn x+y+z=1. Tìm min
$M=\sqrt{x^{2}+xy+y^{2}}+\sqrt{y^{2}+yz+z^{2}}+\sqrt{z^{2}+zx+x^{2}}$
Ta có:$\sqrt{x^2+xy+y^2}=\sqrt{\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2}\geq \sqrt{\frac{3}{4}(x+y)^2}=\frac{\sqrt{3}}{2}(x+y)$$\sqrt{x^2+xy+y^2}=\sqrt{\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2}\geq \sqrt{\frac{3}{4}(x+y)^2}=\frac{\sqrt{3}}{2}(x+y)$.DBXR khi x=y
CMTT:$\sqrt{y^2+yz+z^2}\geq \frac{\sqrt{3}}{2}(y+z)$DBXR khi z=y
$\sqrt{x^2+xz+z^2}\geq \frac{\sqrt{3}}{2}(x+z)$DBXR khi z=x
$\Rightarrow M\geq \frac{\sqrt{3}}{2}(2(x+y+z))=\sqrt{3}$.DBXR khi x=y=z=$\frac{1}{3}$
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