If x,y,z are unequal and if x+y+z=1, the expression \frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}

compyac

compyac

Answered question

2022-04-24

If x,y,z are unequal and if x+y+z=1, the expression (1+x)(1+y)(1+z)(1x)(1y)(1z) is greater than (integer)?

Answer & Explanation

gonzakunti2

gonzakunti2

Beginner2022-04-25Added 16 answers

Step 1
This solution assumes that the terms are non-negative. It doesn't require that the terms are distinct.
With x=y=z=13, we guess that the answer is 8. Let's verify it.
We normalize the inequality by replacing 1=x+y+Z
WTS
(2x+y+z)(2y+z+y)(2z+x+y)8(x+y)(y+z)(z+x)
Can you prove this by applying AM-GM on each term of the product on the LHS

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