system of equations solving for positive $a,b,c$

Question

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?

Answer 1

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: $$2bc(1-(b+c))=(1-(b+c))(b+c)+bc $$ $$\Rightarrow 2bc-2b^2c-2bc^2=b+c-b^2-c^2-2bc+bc $$ $$\Rightarrow 2b^2c+2bc^2+b+c-b^2-c^2-3bc=0 \qquad s.t. \quad 0 \leq b,c \leq 1 $$

Answer 2

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.

Answer 3

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.

Answer 4

I get that there are no solutions to the equations. Here is my reasoning:

Since $(1-a)(1-b)(1-c)=abc$, substituting $c = 1-a-b$ we get $(1-a)(1-b)(a+b)=ab(1-a-b)$ .

Multiplying, $(1-a-b+ab)(a+b) = ab - a^2b-ab^2$ or $a+b-a^2-2ab-b^2+a^2b+ab^2= ab - a^2b-ab^2$ or $a+b-a^2-b^2-3ab+2a^2b+2ab^2 = 0$.

Write this as a quadratic in $b$.

$\begin{align} 0 &=b^2(2a-1)+b(2a^2-3a+1) -a^2+a\\ &=b^2(2a-1)+b(2a-1)(a-1)-a(a-1)\\ \end{align} $.

The discriminant is

$\begin{align} D &= d^2\\ &= ((2a-1)(a-1))^2+4a(a-1)(2a-1)\\ &= (2a-1)(a-1)((2a-1)(a-1)+4a)\\ &= (2a-1)(a-1)(2a^2+a+1)\\ \end{align} $

Since $0 < a < 1$ (by assumption), $a-1<0$ so if $D > 0$ we must have $2a-1 < 0$ or $a < 1/2$.

Since $(2a-1)(a-1) = 2a^2-3a+1 < 2a^2+a+1 $, $0 < ((2a-1)(a-1))^2 < d^2 < (2a^2+a+1)^2$ or $(1-2a)(1-a)=2a^2-3a+1 < d < 2a^2+a+1$

The roots are

$\begin{align} b & = \dfrac{-(2a-1)(a-1)\pm d}{2(2a-1)}\\ & = \dfrac{(1-2a)(1-a)\mp d}{2(1-2a)}\\ & = \dfrac{1-a}{2}\mp\dfrac{ d}{2(1-2a)}\\ \end{align} $

Let $r = \dfrac{ d}{2(1-2a)}$, so the roots are $b_1 = \dfrac{1-a}{2}+r$ and $b_2 = \dfrac{1-a}{2}-r$ .

Since $d > (1-2a)(1-a)$, $r > \dfrac{ (1-2a)(1-a)}{2(1-2a)} = \dfrac{ 1-a}{2} $, so $b_2 < 0$ and connot be a solution.

The last thing we need to see is if we can choose $a$ so that $ b_1 < 1$.

This means that $\dfrac{1-a}{2}+\dfrac{ d}{2(1-2a)}< 1 $ or $(1-a)(1-2a)+ d< 2(1-2a)$ or

$\begin{align} d &< 2(1-2a)-(1-a)(1-2a)\\ &=2-4a-(1-3a+2a^2)\\ &=1-a-2a^2\\ &=(1-2a)(1+a)\\ \end{align} $.

Since $d^2 = (2a-1)(a-1)(2a^2+a+1) = (1-2a)(1-a)(2a^2+a+1) $, we want $(1-2a)(1-a)(2a^2+a+1) < ((1-2a)(1+a))^2 $ or $(1-a)(2a^2+a+1) < (1-2a)(1+a)^2 =(1-2a)(1+2a+a^2) $ or $1+a^2-2a^3 < 1-2a^2-2a^3$ or $3a^2 < 0$.

But this can never hold, so there are no solutions satisfying the equations.

Answer 5

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 $p>0$.