Let \(x,y,z\geq-1\) be real numbers satisfying \(xy+yz+zx+xyz=2\). Find the minimum value of \((1+x)(1+y)(1+z)\) and prove that it is the minimum.

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TopNewestThe minimum value of (1 +

x)(1 +y)(1 +z) would be whenx,y, orzare minimum. However, to satisfy the first equation,x=y=z= -1, because it would yield the equation as 2. Thus, the minimum value of (1 +x)(1 +y)(1 +z) is 0Log in to reply

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