Find all triples of non-negative real numbers (a,b,c)

Question

Find all triples of non-negative real numbers (a,b,c) which satisfy the following set of equations: $ \begin{cases} a^2+ab=c\\ b^2+bc=a\\ c^2+ca=b\\ \end{cases}$

My Attemp:

Solution: The only solutions in non-negative numbers of the system of equations

$(1) \;\; a^2 + ab = c$, $(2) \;\; b^2 + bc = a$, $(3) \;\; c^2 + ca = b$,

are $a=b=c=0$ and ${\textstyle a=b=c=\frac{1}{2}}$.

Proof: Clearly $a=b=c=0$ is a solution of the system of equations (1)-(3).

Next assume $(a,b,c) \neq (0,0,0,)$. According to equations (1)-(3) $abc=0$ iff $a=b=c=0$, implying $a,b,c>0$.

Multiplying the equations (1)-(3), the result is

$abc(a + b)(a + c)(b + c) = abc$,

yielding (since $abc>0$)

$(4) \;\; (a + b)(a + c)(b + c) = 1$.

By adding the equations (1)-(3), we obtain

$a^2 + b^2 + c^2 + ab + ac + bc = a + b + c$,

or alternatively

$(5) \;\; (a + b)^2 + (a + c)^2 + (b + c)^2 = 2(a + b + c)$.

Next set $(x,y,z) = (a+b,a+c,b+c)$. Then by equations (4) and (5)

$(6) \;\; xyz=1$,

$(7) \;\; x^2 + y^2 + z^2 = x + y + z$.

By the AG-inequality and equation (6) we obtain

$x + y + z \geq \sqrt[3]{xyz} = \sqrt[3]{1}$,

i.e.

$(8) \;\; x + y + z \geq 3$.

Moreover by (7)

$(x - 1)^2 + (y - 1)^2 + (z - 1)^2 = 3 - (x + y + z) \geq 0$,

yielding

$(9) \;\; x + y + z \leq 3$.

Combining the inequalities (8) and (9) we obtain

$(10) \;\; x + y + z = 3$.

Hence by (7) and (10)

$3^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) = 3 + 2(xy + xz + yz)$,

which means

$(11) \;\; xy + xz + yz = 3$.

By combining equations (6), (10) and (11) we find that

$(x - 1)(y - 1)(z - 1) = xyz - (xy + xz + yz) + (x + y + z) - 1 = 1 - 3 + 3 - 1$,

i.e.

$(12) \;\; (x - 1)(y - 1)(z - 1) = 0$.

Therefore $x=1$ by (12) (we may WLOG assume $x=1$ since $x,y,z$ are symmetric in equations (6),(10),(11)). Then $yz=1$ and $y + z = 2$ by equations (6) and (10) respectively, yielding

${\textstyle y + \frac{1}{y} = 2}$

$y^2 + 1 = 2y$

$(y - 1)^2 = 0$

$y=1$.

Hence $z=1$ by equation (6). Thus $x=y=z$, yielding $a+b=a+c=b+c$, i.e. $a=b=c$. Consequently by equations (1)-(3) we have

$2a^2 = a$,

which only solution is ${\textstyle a = \frac{1}{2}}$ (since $a>0$). Thus we obtain ${\textstyle a=b=c=\frac{1}{2}}$.

Conclusion: The only solutions of the system of equations (1)-(3) are $a=b=c=0$ and ${\textstyle a=b=c=\frac{1}{2}}$. q.e.d.

Comparing my attempt to the solution below. Did I miss something?

Answer 1

Crucially, at the stage where you had $$a(a+b)+b(b+c)+c(c+a)=a+b+c$$ your assumption that $a(a+b)=a$ and so on was unjustified. If $x+y+z=a+b+c$ then you cannot assume $x=a$ etc.

You might also like to bear in mind that once you have $a(a+b)=a$ there are two possibilities, either $a+b=1$ which you had and also $a=0$.

Answer 2

A quicker way of obtaining the answer

If any of $a,b,c$ is $0$ then they are all $0$. So, suppose $abc>0$ and let $s=a+b+c$. The equations can then be written as $$1+\frac{1}{a}=\frac{s}{c}$$$$ 1+\frac{1}{b}=\frac{s}{a}$$$$1+\frac{1}{c}=\frac{s}{b}.$$

Multiply the equations by $s^2,s,1$, respectively, and add. $$s^2+s+1=\frac{1}{c}(s^3-1)=\frac{1}{c}(s-1)(s^2+s+1).$$ No real $s$ satisfies $s^2+s+1=0$ and so $c=s-1$. Similarly, $a=b=c=s-1$ and the only solution is $a=b=c=\frac{1}{2}.$