Calculate the minimum value of $\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$.

By AM-GM $$\sum_{cyc}\left|\frac{a^2-bc}{b-c}\right|=\sqrt{\left(\left|\sum\limits_{cyc}\frac{a^2-bc}{b-c}\right|\right)^2}=$$ $$=\sqrt{\left(\sum\limits_{cyc}\frac{a^2-bc}{b-c}\right)^2-2\sum_{cyc}\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}+2\sum_{cyc}\left|\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}\right|}\geq$$ $$\geq\sqrt{-2\sum_{cyc}\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}-2\sum_{cyc}\frac{(a^2-bc)(b^2-ac)}{(b-c)(c-a)}}=$$ $$=\sqrt{4(a+b+c)^2}=2(a+b+c)\geq6.$$ Now, prove that we got an infimum and the minimum does not exist.

We have $$ \frac{bc-a^2}{c-b}+\frac{|b^2-ca|}{c-a}+\frac{c^2-ab}{b-a}\geq \frac{bc-a^2}{c-b}-\frac{|b^2-ca|}{c-a}+\frac{c^2-ab}{b-a} $$ with equality only if $b=1$ and $c=\frac{1}{a}$. Now insert $b$ and $c$ to obtain $$ \frac{\frac{1}{a}-a^2}{\frac{1}{a}-1}+\frac{\frac{1}{a^2}-a}{1-a}=\frac{1-a^3}{1-a}+\frac{1-a^3}{a^2-a^3} $$ Differentiation suggest a minimum at $a=1$ but we are not allowed to set $a=1$ however we can take the limit $a\to 1^-$. Do it and use L' Hosptial. $$ \lim_{a\to 1^-}\frac{-3a^2}{-1}+\frac{-3a^2}{2a-3a^2}=6. $$ So, minimum value 6 is obtained when we set $b=1$ and take limits $a\to 1^-$ and $c\to 1^+$ with $c=\frac{1}{a}$.

EDIT: Adding more stuff for clarity

There can only be equality in the inequality if $|b^2-ca|=-|b^2-ca|$ which means equality is when $b^2=ca$, i.e. both sides are zero. Multiply by $b$ and we have $b^3=cab=1$ which gives $b=1$.

Since $b=1$ we get $ac=1$. Choose $a=0.5$ and $c=2$ then we have $\frac{2-0.25}{2-0.5}+\frac{4-0.5}{1-0.5}=\frac{49}{6}>\frac{36}{6}=6$.

Clearly $a<1<c$ otherwise we get division by zero.

Next define $P(a)=\frac{1-a^3}{1-a}+\frac{1-a^3}{a^2-a^3}=a^2+a+2+\frac{1}{a}+\frac{1}{a^2}$ which has extremum at $a=1$. Since we cannot choose $a=1$ we instead take the limit $\lim_{a\to 1^-}P(a)=6$ which is OK since the function $P(a)$ is continuous for $0<a<1$. And we have shown that $P(0.5)>6$ so therefore, because of continuity, we do not need to take second derivative of $P(a)$.

What if $|b^2-ca|<-|b^2-ca|$? Well, it cannot be true since a positive number is larger than a negative number. We have now proven global infimum is 6.