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

Question

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 1

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.

Answer 2

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.

Answer 3

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.