Calculate the minimum value of $\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$. 
Given positives $a, b, c$ such that $abc = 1$, if possible, calculate the minimum value of $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$$

Without loss of generalisation, assume that $a \le b \le c$.
We have that $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right| \ge \frac{c^2 - ba}{b - a} + \frac{a^2 - bc}{b - c}$$
$$ = \frac{(c + a)(a^2 + b^2 + c^2 - bc - ca - ab)}{(c - b)(b - a)} \ (1)$$
Let $c' = b - a, a' = c - b \iff c = a' + b, a = b - c'$, $(1)$ becomes $$\frac{(2b - c' + a')(c'^2 + c'a' + a'^2)}{c'a'}$$
and $(b - c')b(b + a') = b^3 - (c' - a')b^2 - c'a'b = 1$
$$\iff (2b - c' + a')b^2 = b^3 + c'a'b + 1 \iff 2b - c' + a' = \frac{b^3 + c'a'b + 1}{b^2}$$
Another idea I had was that $\left|\dfrac{a^2 - bc}{b - c}\right| + \left|\dfrac{b^2 - ca}{c - a}\right| + \left|\dfrac{c^2 - ab}{a - b}\right|$
$$ = \frac{1}{2}\sum_{\text{cyc}}\left(|c - a|\left|\frac{2(b^2 - ca)}{(c - a)^2}\right|\right) = \frac{1}{2}\sum_{\text{cyc}}\left(|c - a|\left|\frac{2b^2 - c^2 - a^2}{(c - a)^2} + 1\right|\right)$$
$$ = \frac{1}{2}\left[(c - b)\left(\left|\frac{2(b^2 - a^2)}{(b - c)^2} + \frac{c^2 - b^2}{(b - c)^2} - 1\right| + \left|\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + 1\right|\right)\right.$$
$$\left. + (b - a)\left(\left|\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + 1\right| + \left|\frac{b^2 - a^2}{(a - b)^2} + \frac{2(c^2 - b^2)}{(a - b)^2} + 1\right|\right)\right]$$
$$ \ge \frac{1}{2}\left[(c - b)\left(\frac{2(b^2 - a^2)}{(b - c)^2} + \frac{c^2 - b^2}{(b - c)^2} + \frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2}\right)\right.$$
$$\left. + (b - a)\left(\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + \frac{b^2 - a^2}{(a - b)^2} + \frac{2(c^2 - b^2)}{(a - b)^2} + 2\right)\right]$$
$$ = \frac{1}{2}\left[(c^2 - b^2)\left(\frac{1}{c - b} - \frac{1}{c - a} + \frac{2}{b - a}\right) + (b^2 - a^2)\left(\frac{2}{c - b} + \frac{1}{c - a} + \frac{1}{b - a}\right)\right] + (b - a)$$
$$ = \frac{1}{2}\left(\frac{b^2 + c^2 - 2a^2}{c - b} + \frac{2b^2 - c^2 - a^2}{c - a} + \frac{2c^2 - a^2 - b^2}{b - a}\right) + (b - a)$$
There must have been something wrong, but that's all I have for now.

Answer to the question: $\lim_{h \to 0}6+h$
The given expression$$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right| $$
upon condition that $abc=1$
This can be rewritten as
$$\left|\frac{a^2 - \frac{1}{a}}{b - c}\right| + \left|\frac{b^2 - \frac{1}{b}}{c - a}\right| + \left|\frac{c^2 - \frac{1}{c}}{a - b}\right| $$ which is further reducable to $$\left|\frac{a-1}{b - c}(a+\frac{1}{a}+1)\right| + \left|\frac{b-1}{c-a}(b+\frac{1}{b}+1)\right| + \left|\frac{c-1}{a-b}(c+\frac{1}{c}+1)\right| $$
if $a,b,c >0$
this reduces to $$3\big(\left|\frac{a-1}{b - c}\right| + \left|\frac{b-1}{ c-a}\right| + \left|\frac{c-1}{a-b}\right|\big) $$
Note: The reduction for $a+\frac{1}{a} +1>3$ and not $\geq 3$ because doing so would imply all variables equal $1$
The least value for this expression I could yield using $a$ fixed at 1 and manipulating $b,c$ was 6
Since there's a modulus operation, the least we can get out of |x| is 0, so set $a=1$, by doing that we are left with
$$\left|\frac{b-1}{ c-1}\right| + \left|\frac{c-1}{b-1}\right|\ $$
which is definitely $$> 2$$ (remember $b$ and $c$ cannot be equal)
Hence the answer.
