<s>[INST] If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$ If $\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0$ with unequal $a,b,c$, Prove that $\dfrac{a}{(b-c)^2}+\dfrac{b}{(c-a)^2}+\dfrac{c}{(a-b)^2}=0$  
I could not approach the problem at all though I think I could have done something by using Cauchy-Schwarz inequality, but could not pull this together. Please help.
 [/INST] Let 
$$
T_1=\sum_{cyc} \frac{a}{b-c}
$$
By hypothesis we have $T_1=0$, but if we put
$T_2=T_1(ab+ac+bc-a^2-b^2-c^2)$, we have 
$$
T_2=\sum_{cyc} \frac{a(ab+ac+bc-a^2-b^2-c^2)}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a^2b}{b-c}+
\sum_{cyc} \frac{a^2c}{b-c}+
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}-
\sum_{cyc} \frac{ab^2}{b-c}-
\sum_{cyc} \frac{ac^2}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a^2b}{b-c}+
\sum_{cyc} \frac{a^2b}{c-b}+
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}-
\sum_{cyc} \frac{ab^2}{b-c}-
\sum_{cyc} \frac{ab^2}{c-b}
$$
$$
T_2=
\sum_{cyc} \frac{abc}{b-c}-
\sum_{cyc} \frac{a^3}{b-c}
=\sum_{cyc} \frac{a(bc-a^2)}{b-c}
$$
$$
T_2=\sum_{cyc} \frac{a(bc-a^2)}{b-c}
+\sum_{cyc} \frac{-a^2b}{b-c}
+\sum_{cyc} \frac{-a^2c}{b-c}
$$
and hence
$$
T_2=\sum_{cyc} \frac{a(b-a)(c-a)}{b-c}=
-\sum_{cyc} \frac{a(a-b)(c-a)}{b-c}
$$
So if we put $T_3=\frac{T_2}{(a-b)(b-c)(c-a)}$, we see that
$$
T_3=-\sum_{cyc} \frac{a}{(b-c)^2}
$$
and we are done.
</s>