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

0<x;y;z<1 (1-x)(1-y)(1-z)=xyz Find: Ω = min (((1-x)/(xy)) + ((1-y)/(yz)) + ((1-z)/(zx)))

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

Answered by aleks041103 last updated on 24/Dec/21

1−x−y−z+xy+xz+yz=2xyz ((1-x)/(xy)) + ((1-y)/(yz)) + ((1-z)/(zx))= =((z−zx+x−xy+y−yz)/(xyz))= =((1−2xyz)/(xyz))=(1/(xyz))−2 We need to find max of xyz: (1−x)(1−y)(1−z)=xyz (1/z)−1=((xy)/((1−x)(1−y))) ⇒z=(1/(1+((xy)/((1−x)(1−y)))))=(((1−x)(1−y))/((1−x)(1−y)+xy))= =(((1−x)(1−y)+xy)/((1−x)(1−y)+xy))−((xy)/((1−x)(1−y)+xy))= =1+((xy)/(1−x−y+2xy))=z(x,y) ⇒f(x,y,z)=xyz(x,y)= =xy+((x^2 y^2 )/(1−x−y+2xy)) f_x =(1+((2xy)/(1−x−y+2xy))−((x^2 y(2y−1))/((1−x−y+2xy)^2 )))y=0 f_y =(1+((2xy)/(1−x−y+2xy))+((xy^2 (2x−1))/((1−x−y+2xy)^2 )))x=0 ⇒(1−x−y+2xy)^2 +2xy(1−x−y+2xy)+x^2 y(2x−1)=0 ⇒(1−x−y+2xy)^2 +2xy(1−x−y+2xy)+y^2 x(2y−1)=0 ⇒x^2 (2x−1)y=y^2 x(2y−1)⇒x(2x−1)=y(2y−1)=a ⇒x,y are solutions to 2p^2 −p−a=0 x=y (1−2x+2x^2 )^2 +2x^2 (1−2x+2x^2 )+x^3 (2x−1)=0 ...

$$\mathrm{1}−{x}−{y}−{z}+{xy}+{xz}+{yz}=\mathrm{2}{xyz} \\ $$$$\frac{\mathrm{1}-\mathrm{x}}{\mathrm{xy}}\:+\:\frac{\mathrm{1}-\mathrm{y}}{\mathrm{yz}}\:+\:\frac{\mathrm{1}-\mathrm{z}}{\mathrm{zx}}= \\ $$$$=\frac{{z}−{zx}+{x}−{xy}+{y}−{yz}}{{xyz}}= \\ $$$$=\frac{\mathrm{1}−\mathrm{2}{xyz}}{{xyz}}=\frac{\mathrm{1}}{{xyz}}−\mathrm{2} \\ $$$${We}\:{need}\:{to}\:{find}\:{max}\:{of}\:{xyz}: \\ $$$$\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right)={xyz} \\ $$$$\frac{\mathrm{1}}{{z}}−\mathrm{1}=\frac{{xy}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)} \\ $$$$\Rightarrow{z}=\frac{\mathrm{1}}{\mathrm{1}+\frac{{xy}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)}}=\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)+{xy}}= \\ $$$$=\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)+{xy}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)+{xy}}−\frac{{xy}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)+{xy}}= \\ $$$$=\mathrm{1}+\frac{{xy}}{\mathrm{1}−{x}−{y}+\mathrm{2}{xy}}={z}\left({x},{y}\right) \\ $$$$\Rightarrow{f}\left({x},{y},{z}\right)={xyz}\left({x},{y}\right)= \\ $$$$={xy}+\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{\mathrm{1}−{x}−{y}+\mathrm{2}{xy}} \\ $$$${f}_{{x}} =\left(\mathrm{1}+\frac{\mathrm{2}{xy}}{\mathrm{1}−{x}−{y}+\mathrm{2}{xy}}−\frac{{x}^{\mathrm{2}} {y}\left(\mathrm{2}{y}−\mathrm{1}\right)}{\left(\mathrm{1}−{x}−{y}+\mathrm{2}{xy}\right)^{\mathrm{2}} }\right){y}=\mathrm{0} \\ $$$${f}_{{y}} =\left(\mathrm{1}+\frac{\mathrm{2}{xy}}{\mathrm{1}−{x}−{y}+\mathrm{2}{xy}}+\frac{{xy}^{\mathrm{2}} \left(\mathrm{2}{x}−\mathrm{1}\right)}{\left(\mathrm{1}−{x}−{y}+\mathrm{2}{xy}\right)^{\mathrm{2}} }\right){x}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}−{x}−{y}+\mathrm{2}{xy}\right)^{\mathrm{2}} +\mathrm{2}{xy}\left(\mathrm{1}−{x}−{y}+\mathrm{2}{xy}\right)+{x}^{\mathrm{2}} {y}\left(\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}−{x}−{y}+\mathrm{2}{xy}\right)^{\mathrm{2}} +\mathrm{2}{xy}\left(\mathrm{1}−{x}−{y}+\mathrm{2}{xy}\right)+{y}^{\mathrm{2}} {x}\left(\mathrm{2}{y}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{2}{x}−\mathrm{1}\right){y}={y}^{\mathrm{2}} {x}\left(\mathrm{2}{y}−\mathrm{1}\right)\Rightarrow{x}\left(\mathrm{2}{x}−\mathrm{1}\right)={y}\left(\mathrm{2}{y}−\mathrm{1}\right)={a} \\ $$$$\Rightarrow{x},{y}\:{are}\:{solutions}\:{to}\:\mathrm{2}{p}^{\mathrm{2}} −{p}−{a}=\mathrm{0} \\ $$$${x}={y} \\ $$$$\left(\mathrm{1}−\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} \right)+{x}^{\mathrm{3}} \left(\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$... \\ $$

Commented by HongKing last updated on 28/Dec/21

thank you dear Sir cool

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

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