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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 byHongKing last updated on 28/Dec/21

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

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