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Question Number 166830 by HongKing last updated on 28/Feb/22

Answered by nurtani last updated on 01/Mar/22

$$(x+\frac{1}{y})(y+\frac{1}{z})(z+\frac{1}{x})=(xy+\frac{x}{z}+1+\frac{1}{yz})(z+\frac{1}{x})=xyz+x+z+\frac{1}{y}+y+\frac{1}{z}+\frac{1}{x}+\frac{1}{xyz}=\left(4\right)\left(1\right)\left(\frac{7}{3}\right)$$$$\iff xyz+\frac{1}{xyz}+(x+\frac{1}{y})+(y+\frac{1}{z})+(z+\frac{1}{x})=\left(4\right)\left(1\right)\left(\frac{7}{3}\right)=\frac{28}{3}$$$$\iff xyz+\frac{1}{xyz}+4+1+\frac{7}{3}=\frac{28}{3}$$$$\iff xyz+\frac{1}{xyz}+\frac{22}{3}=\frac{28}{3}\iff xyz+\frac{1}{xyz}=\frac{6}{3}=2$$$$\iff x^2{y}^2{z}^2+1=2xyz$$$$\iff {\left(xyz\right)}^2-2xyz+1=0$$$$let:\phi =xyz\Rightarrow {\phi }^2-2\phi +1=0$$$$\Rightarrow {(\phi -1)}^2=0\Rightarrow \phi =1$$$$\therefore \phi =xyz=1$$

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