If x,y,z are unequal and if $x+y+z=1$, the expression ${(1+x)(1+y)(1+z)\over (1-x)(1-y)(1-z)}$ is greater than(integer)?

Question

If $x, y, z$ are unequal and if $x+y+z=1$, the expression $${(1+x)(1+y)(1+z)\over (1-x)(1-y)(1-z)}$$ is greater than (integer)?

I thought of doing this question by the inequality approach like AM, GM, HM, Cauchy-Swartz etc and just got to $(1+x)(1+y)(1+z)\ge8(x+y)(y+z)(z+x)$ but I am not able to relate it to $(1-x)(1-y)(1-z)$.
And is there any more method you can suggest for this question or likewise types?

Answer 1

This solution assumes that the terms are non-negative. It doesn't require that the terms are distinct.

With $ x = y = z = 1/3$, we guess that the answer is 8. Let's verify it.

We normalize the inequality by replacing $ 1 = x+y+z$. WTS

$$ ( 2x+y+z)(2y+z+y)(2z+x+y) \geq 8 (x+y)(y+z)(z+x)$$

Can you prove this by applying AM-GM on each term of the product on the LHS?

$2x + y + z = (x+y) + (x+z) \geq 2 \sqrt{ (x+y)(x+z) }$.
We have analogous inequalities for the other terms.
Multiply them together, and we get the inequality.