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?
 A: 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:
$$2bc(1-(b+c))=(1-(b+c))(b+c)+bc
$$
$$\Rightarrow 2bc-2b^2c-2bc^2=b+c-b^2-c^2-2bc+bc
$$
$$\Rightarrow 2b^2c+2bc^2+b+c-b^2-c^2-3bc=0 \qquad s.t. \quad 0 \leq b,c \leq 1
$$
A: If any two of $a,b,c$ are zero, with the third equal to $1$, both equations are satisfied: so $0<a,b,c<1$ are not correct "strict" bounds. 
So we have the following three solutions for $(a, b, c)$:


*

*$(1, 0, 0)$

*$(0, 1, 0)$

*$(0, 0, 1)$


You have a degree three equation: $(1 - a)(1-b)(1 - c)=abc$, and a linear (degree 1) equation: $a + b + c = 1,\;$ so we can rest with our three solutions.
A: From $(1-a)(1-b)(1-c)=abc$ you cannot deduce $(1-a)=1$ and so on.  One set of solutions is to make both sides zero-let $a=1,b=0,c=0$ or any permutation.  The symmetry means that any permutation of a solution will also be a solution.  Given one cubic and one linear equation you expect three solutions and we have that many.
A: There are no positive $a$, $b$, $c$ that satisfy your equations. 
Proof. Let 
$$\sigma_1:=a+b+c,\quad \sigma_2:=ab+bc+ca,\quad \sigma_3:=abc\ .$$
Then your two equations are equivalent to
$$ \sigma_1=1\qquad\wedge\qquad \sigma_2=2\sigma_3\ .$$ It follows that the three numbers $a$, $b$, $c$ would have to be the solutions of an equation of the form
$$f(t):=t^3-t^2+2pt-p=0$$
for some $p>0$. To enable three real zeros of $f$ the derivative $f'(t)=3t^2-2t+2p$ would have to have two real zeros
$$t_i={2\pm\sqrt{4-24 p}\over 6}\qquad(i=1,\>2)\ ,$$
which requires $0<p\leq{1\over6}$. In addition it would be necessary that the the values $f(t_1)$ and $f(t_2)$ have opposite sign. Computation gives
$$f(t_1)\>f(t_2)={1\over27}p(4-13p+32p^2)\ ,$$
and this cannot be negative when $p>0$.
