<s>[INST] 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.
 [/INST] 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:
$$\left\{(0,0,0),\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\right\}$$
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