<s>[INST] Proof verification: $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ The question states:

Let $a, b, c$ be real numbers such that $$\frac{1}{(bc-a^2)}+\frac{1}{(ca-b^2)}+\frac{1}{(ab-c^2)}=0$$ Prove that $$\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$$

Now this can be solved with algebraic manipulation, but I want to check if my solution is valid too:
By Cauchy-Schwarz,
$$\left (\frac{a^2}{(bc-a^2)^3}+\frac{b^2}{(ca-b^2)^3}+\frac{c^2}{(ab-c^2)^3}\right )\left (\frac{1}{(bc-a^2)}+\frac{1}{(ca-b^2)}+\frac{1}{(ab-c^2)}\right ) \ge \left (\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}\right )^2 = P^2$$
Hence, $0 \ge P^2 \implies P = 0$.
 [/INST] C-S it's the following:
$$\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2\geq\left(\sum_{i=1}^na_ib_i\right)^2$$ for any reals $a_i$ and $b_i$,
which says that your solution is wrong because $bc-a^2$ may be negative and can not be square of real number.  You can not use C-S so.
By the way, $$\sum_{cyc}\frac{1}{ab-c^2}=\frac{(ab+ac+bc)\sum\limits_{cyc}(a^2-bc)}{\prod\limits_{cyc}(a^2-bc)}$$ and
$$\sum_{cyc}\frac{a}{(bc-a^2)^2}=\frac{(a+b+c)(ab+ac+bc)\sum\limits_{cyc}(a^2-bc)\sum\limits_{cyc}(a^2b^2-a^2bc)}{\prod\limits_{cyc}(a^2-bc)^2},$$ which gives the solution.
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