<s>[INST] $x\over(1-x)$ $y\over(1-y)$ $z\over(1-z)$ >= 8 when $x ,y ,z $ are positive proper fractions and $x+y+z = 2$ Q. Prove that $x\over(1-x)$ $y\over(1-y)$ $z\over(1-z)$ $\geq$ 8 when $x ,y ,z $ are positive proper fractions and $x+y+z = 2$
What I did: From A.M. G.M. inequality,
$(x+y+z)\over3$ $\geq$ $(xyz)^{1/3}$
or, $xyz \leq$ $8\over27$ (Putting $x+y+z = 2$ and cubing both sides) -----------1.
And, $(1-x)+(1-y)+(1-z)\over3 $$\geq [(1-x)(1-y)(1-z)]^{1/3}$
or, $1/3 \geq [(1-x)(1-y)(1-z)]^{1/3}$ (Putting $x+y+z = 2$)
or, $1\over(1-x)(1-y)(1-z)$ $\geq$ $1\over27$ ----------------------2.
But we cannot multiply 1. by 2. Since the inequality signes are different.
Am I missing something?
 [/INST] Note that $1 = (x+y+z)/2$. The inequality becomes
$$  \frac{xyz}{(y+z-x)(x+z-y)(x+y-z)} \geq 1$$
which is equivalent to 
$$ xyz \geq (y+z-x)(x+z-y)(x+y-z)\  \ \  (1)$$
To see why this is true, apply AM-GM inequality three times:
$$ x = \frac{x+y-z+x+z-y}{2} \geq \sqrt{(x+y-z)(z+x-y)}$$
and the other two for $y,z$. Now multiply all three and you'll get $(1)$.
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