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Question Number 161900 by HongKing last updated on 23/Dec/21

$$0<\mathrm{x};\mathrm{y};\mathrm{z}<1$$$$(1-\mathrm{x})(1-\mathrm{y})(1-\mathrm{z})=\mathrm{xyz}$$$$\mathrm{Find}:$$$$\mathrm{\Omega }=\min (\frac{1-\mathrm{x}}{\mathrm{xy}}+\frac{1-\mathrm{y}}{\mathrm{yz}}+\frac{1-\mathrm{z}}{\mathrm{zx}})$$

Answered by aleks041103 last updated on 24/Dec/21

$$1-x-y-z+xy+xz+yz=2xyz$$$$\frac{1-\mathrm{x}}{\mathrm{xy}}+\frac{1-\mathrm{y}}{\mathrm{yz}}+\frac{1-\mathrm{z}}{\mathrm{zx}}=$$$$=\frac{z-zx+x-xy+y-yz}{xyz}=$$$$=\frac{1-2xyz}{xyz}=\frac{1}{xyz}-2$$$$Weneedtofindmaxofxyz:$$$$(1-x)(1-y)(1-z)=xyz$$$$\frac{1}{z}-1=\frac{xy}{(1-x)(1-y)}$$$$\Rightarrow z=\frac{1}{1+\frac{xy}{(1-x)(1-y)}}=\frac{(1-x)(1-y)}{(1-x)(1-y)+xy}=$$$$=\frac{(1-x)(1-y)+xy}{(1-x)(1-y)+xy}-\frac{xy}{(1-x)(1-y)+xy}=$$$$=1+\frac{xy}{1-x-y+2xy}=z(x,y)$$$$\Rightarrow f(x,y,z)=xyz(x,y)=$$$$=xy+\frac{x^2{y}^2}{1-x-y+2xy}$$$$f_x=(1+\frac{2xy}{1-x-y+2xy}-\frac{x^{2}y(2y-1)}{{(1-x-y+2xy)}^2})y=0$$$$f_y=(1+\frac{2xy}{1-x-y+2xy}+\frac{{xy}^2(2x-1)}{{(1-x-y+2xy)}^2})x=0$$$$\Rightarrow {(1-x-y+2xy)}^2+2xy(1-x-y+2xy)+x^{2}y(2x-1)=0$$$$\Rightarrow {(1-x-y+2xy)}^2+2xy(1-x-y+2xy)+y^{2}x(2y-1)=0$$$$\Rightarrow x^2(2x-1)y=y^{2}x(2y-1)\Rightarrow x(2x-1)=y(2y-1)=a$$$$\Rightarrow x,yaresolutionsto2p^2-p-a=0$$$$x=y$$$${(1-2x+2x^2)}^2+2x^2(1-2x+2x^2)+x^3(2x-1)=0$$$$...$$

Commented by HongKing last updated on 28/Dec/21

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{Sir}\mathrm{cool}$$

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