Prove that $a+b+c\le 0$

Question

$a,b,c$ are integer numbers different from zero and $$\frac{a}{b+c^2}=\frac{a+c^2}{b}$$ Prove that $a+b+c\le 0$. I know how to prove that $a+b<0$ but do not know what about $c$.

Answer 1

Since $c \in \mathbb{Z}$, we have $c \le |c|\le c^2$

$$0=(a+b)c^2+c^4$$

$$a+b+c^2=0$$

$$a+b+c \le a+b+c^2=0$$

Answer 2

We have $$ab=(a+c^2)(b+c^2)\\ab=ab+c^4+c^2(a+b)$$which means that $$a+b=-c^2\le -c$$or equivalently $$a+b+c\le 0$$

Answer 3

From given equation we get $a+b+c^2 =0$ (since $c\ne 0$), so we have $$a+b+c= c-c^2 = c(1-c) \leq 0$$

If $c>0$ then $1-c\leq 0$ so $c(1-c)\leq 0$ and

if $c<0$ then $1-c>0$ so $c(1-c)<0$