[Toán 9] Mình cần giúp đỡ

M

minhtuyb

Từ gt:[TEX](x+y+z)xyz=1\Rightarrow x+y+z=\frac{1}{xyz}[/TEX]
Có:[TEX](x+y)(y+z)=y^2+xy+yz+xz=y(x+y+z)+xz=y.\frac{1}{xyz}+xz=\frac{1}{xz}+xz\geq ^{Cauchy} 2\sqrt{\frac{1}{xz}.xz}=2<DPCM>[/TEX]
Dấu bằng xảy ra khi [TEX]xz=1[/TEX]
 
You must log in or register to reply here.
Top Bottom