$$\frac{a}{b+c^2}=\frac{a+c^2}{b}$$$$ab=(a+c^2)(b+c^2)$$$$ab=ab+ac^2+bc^2+c^4$$$$c^2(a+b+c^2)=ab-ab=0$$Now,we will only look at the case when $a+b+c^2=0$,because the other case implies $c^2=0\Rightarrow c=0$ which we cannot have,so: $$a+b+c^2=0\Rightarrow c^2=-a-b$$Let $a_1$ and $b_1$ be integers such that $a=-a_1$ and $b=-b_1$ $$c^2=a_1+b_1\Rightarrow c=\sqrt{a_1+b_1}=({a_1+b_1})^\frac{1}{2}\leq{(a_1+b_1)}^1=a_1+b_1$$$$\Rightarrow c\leq a_1+b_1$$Multiplying the last expression by $-1$ we have:$-c\geq -a_1-b_1=a+b$ $<=>$ $a+b+c\leq 0$ Q.E.D