<s>[INST] An inequality with a weird condition $a,b,c$ are distinct real numbers that $$(a^2)(1-b+c)+(b^2)(1-c+a)+(c^2)(1-a+b)=ab+bc+ca.$$ Prove $$\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=1.$$
I have tried two different approaches to this problem but am stuck.

*

*First I started by expanding the bottom of each fraction.
$\frac{1}{(a^2)-2ab+(b^2)}+\frac{1}{(b^2)-2bc+(c^2)}+\frac{1}{(c^2)-2ac+(a^2)}=1$


*I also tried to expand $(a^2)(1-b+c)+(b^2)(1-c-a)+(c^2)(1-a+b)=ab+bc+ca$
$(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b=ab+bc+ca$
$(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca=0$
Then I added one to both sides
$(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca+1=0+1$
Help will be appreciated.
 [/INST] Hint: I'll assume the second factor of term two is $(1-c+a)$ as one would expect in such cyclic expressions. If now one considers the left side $w$ of the thing to be shown equal to $1$ (i.e. the sum of the reciprocal squares of differences), it factors nicely as
$$w=\frac{(a^2+b^2+c^2-ab-ac-bc)^2}{[(a-b)(b-c)(c-a)]^2}.$$
So $w=1$ iff the unsquared top (call that $T$) of it is $\pm$ its unsquared bottom (call that $B$).
But if one expands the assumed equation, after moving the right side over, and compares with the sum $T+B$ or difference $T-B$ there's a match which I think shows what you want. I haven't completely spelled it out, hence called it a hint.
</s>