<s>[INST] system of equations solving for positive $a,b,c$ i need help 
i need to find positive number $a,b,c$ solving this system of equations?

$$(1-a)(1-b)(1-c)=abc$$
$$a+b+c=1$$
I found that $0<a,b,c<1$ and I try to solve it by try $(1-a)=a$, $(1-b)=b$ and $(1-c)=c$ and got that 
$a=0.5,b=0.5,c=0.5$ but it contradicts the second equations.
Can someone help me?
 [/INST] by expanding the left hand side you will have
$$1-(a+b+c)+ab+bc+ac-abc=abc
$$
using the second equality we will have:
$$2abc=ab+bc+ca
$$
And using the substitution method we can replace $a$ by $1-(b+c)$ then we have:
$$2bc(1-(b+c))=(1-(b+c))(b+c)+bc
$$
$$\Rightarrow 2bc-2b^2c-2bc^2=b+c-b^2-c^2-2bc+bc
$$
$$\Rightarrow 2b^2c+2bc^2+b+c-b^2-c^2-3bc=0 \qquad s.t. \quad 0 \leq b,c \leq 1
$$
</s>