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If xyz=(1-x)(1-y)(1-z) Where 0<=x,y, z<=...

If `xyz=(1-x)(1-y)(1-z)` Where `0<=x,y, z<=1`, then the minimum value of `x(1-z)+y (1-x)+ z(1-y)` is

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we will apply the concept of `AM>=GM`
the three terms are x(1-z)+y(1-x)+z(1-y)
=> `(x(1-z)+y(1-x)+z(1-y))/3 >=(x(1-z)y(1-x)z(1-y))^(1/3)`
=>equality holds when all numbers are equal
=>x(1-z)=y(1-x)=z(1-y)
on solving we get to know that `x=y=z=1/2`
=> `x(1-z)+y(1-x)+z(1-y)>=3/4`
thus minimum value of required expression is `3/4`
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