<s>[INST] Prove that $a+b+c\le 0$ $a,b,c$ 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 $c$.
 [/INST] 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 $c(1-c)<0$
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