Question

Answer 1

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Answer 2

\(xyz=1\Leftrightarrow\left\{{}\begin{matrix}xy=\dfrac{1}{z}\\yz=\dfrac{1}{x}\\xz=\dfrac{1}{y}\end{matrix}\right.\)

\(\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)\left(z+\dfrac{1}{z}\right)\\ =\left(xy+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{1}{xy}\right)\left(z+\dfrac{1}{z}\right)\\ =xyz+\dfrac{xy}{z}+\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}+\dfrac{1}{xyz}\\ =2+\dfrac{1}{z^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2}+x^2+y^2+z^2\)

\(\Leftrightarrow P=x^2+y^2+z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+8-2-\dfrac{1}{x^2}-\dfrac{1}{y^2}-\dfrac{1}{z^2}-x^2-y^2-z^2\\ \Leftrightarrow P=6\)