FINAL ROUND

If $b=0$ or $c=0$, then $bc$ is a perfect square. If $a=0$, then from $$\frac{a^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}=\frac{2c}{b+c}(1)$$ it follows that $b=c$, so $bc$ is a perfect square. Therefore we can assume that all $a$, $b$, $c$ are different from $0$. Then we can rewrite (1) as $$\frac{1}{1+\frac{b^2}{a^2}}+\frac{1}{1+\frac{a^2}{c^2}}=\frac{2}{1+\frac{b}{a}\cdot \frac{c}{c}}$$ Set $x=\frac{b}{a}$ and $y=\frac{a}{c}$, we have $$\frac{1}{1+x^2}+\frac{1}{1+y^2}=\frac{2}{1+xy}\iff \frac{2+x^2+y^2}{(1+x^2)(1+y^2)}=\frac{2}{1+xy}\iff$$ $$(2+x^2+y^2)(1+xy)=2(1+x^2)(1+y^2)\iff$$ $$2+x^2+y^2+2xy+x^{3}y+x{y}^3=2+2x^2+2y^2+2x^2{y}^2\iff$$ $$x^{3}y-2x^2{y}^2+x{y}^3-x^2+2xy-y^2=0\iff xy(x-y{)}^2-(x-y{)}^2=0\iff (x-y{)}^2(xy-1)=0.$$ Then, if $x-y=0$, that is $x=y$, then $\frac{b}{a}=\frac{a}{c}$, and $bc=a^2$ is a perfect square. If $xy=1$, that is $\frac{b}{a}\cdot \frac{a}{c}=1\iff \frac{b}{c}=1$, then $b=c$, so $bc$ is a perfect square.