Find all triples of non-negative real numbers $(a,b,c)$ Find all triples of non-negative real numbers $(a,b,c)$ such that:
$a^2+ab=c$
$b^2+bc=a$
$c^2+ca=b$.
This question was problem number of 3 in the RMO(India) Olympiad in 2019 held on $10^{th}$ November. link  .

My attempt-
Assume $a \geq b\geq c$,
$\therefore a^2 \geq b^2$ and $ab\geq bc$.
Adding these two, $a^2+ab\geq b^2+bc$
implies $c \geq a$.
Which can be possible only if  $a=c$, which also implies $a=b=c$.
Substituting in the base equations,
$a^2+a^2=a$
$\therefore (a,b,c)=(0,0,0)$ or $(0.5,0.5,0.5)$
I want to know if my method is correct because it seems a lot different than the one provided in the solutions.
 A: Your method is not full because the system is cyclic and not symmetric.
You can not assume that $a\geq b\geq c$.
We can solve our system by the following way.
We have: if $a=0$ so $c=0$ and from here $b=0$, which gives a solution $(0,0,0)$.
Thus, it's enough to solve our system for $abc\neq0$ and sice
$$a(a+b)=c,$$
$$b(b+c)=a$$ and $$c(c+a)=b,$$ we obtain:
$$abc\prod_{cyc}(a+b)=abc$$ or
$$\prod_{cyc}(a+b)=1.$$
In another hand, from the starting system we obtain:
$$(a-b)(a+b)=(c-a)(b+1),$$
$$(b-c)(b+c)=(a-b)(c+1)$$ and
$$(c-a)(c+a)=(b-c)(a+1).$$
Thus, $a=b$ gives $c=a$ and we got as you got: $a=b=c,$ which gives also a solution $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right).$
Now, let $(a-b)(b-c)(c-a)\neq0.$
Thus, from the last system we obtain:
$$\prod_{cyc}(a-b)\prod_{cyc}(a+b)=\prod_{cyc}(a-b)\prod_{cyc}(a+1)$$ or
$$1=\prod_{cyc}(a+1),$$ which is impossible, which gives the answer:
$$\left\{(0,0,0),\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\right\}$$
A: I am not quite convinced by your method. 
When we have equations that are completely symmetric in three variables, we can assume any partial order we like without loss of generality. 
In this case, however, the equations are only cyclically symmetric in $a,$ $b,$ and $c.$
If you swap $a$ and $b$ you get a different set of equations. 
But if you also mention that a similar result occurs when $b\geq a\geq c,$ then you have covered all possible permutations of the variables in the partial order, and I am convinced. Fortunately, a similar result does follow. 
A: Eliminating $b,c$ from our system we get for $a$ the following equation:
$$2 a^8+a^7+a^6+3 a^5+6 a^4+2 a^3-a^2-a=0$$
and this is $$a (2 a-1) \left(a^6+a^5+a^4+2 a^3+4 a^2+3 a+1\right)=0$$
