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If x,y,z∈R\displaystyle x,y,z\in\mathbb Rx,y,z∈R are positive, solve {x+1yz+z2=3y+1xz=2\displaystyle\begin{cases}x+\frac{1}{yz}+z^2=3\\y+\frac{1}{xz}=2\end{cases}{x+yz1+z2=3y+xz1=2
The solutions are (x1,y1,z1),(x2,y2,z2),…,(xn,yn,zn)\displaystyle (x_1,y_1,z_1), (x_2,y_2,z_2),\ldots, (x_n,y_n,z_n)(x1,y1,z1),(x2,y2,z2),…,(xn,yn,zn).
Find ∑i=1n(xi+yi+zi)\displaystyle\sum_{i=1}^n (x_i+y_i+z_i)i=1∑n(xi+yi+zi).
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