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, $0 \ge P^2 \implies P = 0$.
 A: 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.
A: Your approach is not valid, and by coincidence, you don't arrive at an obvious contradiction because the final answer is zero too.
Let's examine a numerical example.
For $a=1, b=1, c=\frac{-1}{2}$, we get $\frac {1}{bc-a^2}+\frac {1}{ca-b^2}+\frac {1}{ab-c^2}=0$. If your approach was supposed to be correct, in a similar way, we would get:
$$(a^2(bc-a^2)+b^2(ac-b^2)+c^2(ab-c^2))(\frac {1}{bc-a^2}+\frac {1}{ca-b^2}+\frac {1}{ab-c^2})\ge (a+b+c)^2,$$
therefore we would have $a+b+c=0$, which is impossible according to the set of $a=1, b=1, c=\frac{-1}{2}$.
Indeed, the correct inequality we are allowed to use is:
$$(|(a^2(bc-a^2)|+|b^2(ac-b^2)|+|c^2(ab-c^2)|)(|\frac {1}{bc-a^2}|+|\frac {1}{ca-b^2}|+|\frac {1}{ab-c^2}|)\ge (a+b+c)^2,$$
and the same goes for the inequality you used (considering the absolute values).
Similar to Michael Rozenberg's solution, you can see this link.
