If $a,b,c\in\mathbb{R}$ and $a+b+c=1$, then what is the minimum value
of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$.
I used Wolfram Alpha and it says the minimum value is $\dfrac{44}{27}$ for $\left(a,b,c\right)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. Obviously $uvw$ method doesn't help. and since we have absolute value, I think The Buffalo doesn't help.
I wrote $ab+bc+ca$ this way: $abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.
We know that
$$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\le\frac{a+b+c}{3},$$
so
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\Rightarrow ab+bc+ca\ge9abc$$
I don't know if it helps (or is even true).
I think finding the minimum of $\left(1-\left(ab+bc+ca\right)\right)^2+\left(1-abc\right)^2$ has the same procedure. If anyone knows how to find it, it helps a lot (I think).
Best Answer
Let $p = a + b + c = 1$, $q = ab + bc + ca$ and $r = abc$.
Fact 1: $q^2 \ge 3pr$.
(The proof is given at the end.)
Fact 2: $p^2 \ge 3q$.
(Proof: $p^2 - 3q = \frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2] \ge 0$.)
We split into four cases:
$q \ge 0$: By Fact 2, we have $q \le \frac{1}{3}$. By Facts 1-2, we have $r\le \frac{1}{27}$. We have $|1- q| + |1 - r| = 1 - q + 1 - r = 2 - q - r \ge 2 - \frac{1}{3} - \frac{1}{27} = \frac{44}{27}$ with equality if $a = b = c = \frac{1}{3}$.
$q < 0$ and $r > 1$: By Fact 1, we have $q^2 \ge 3$. Thus, $q < -\sqrt{3}$. We have $|1-q| + |1 - r| = 1 - q + r - 1 = -q + r > \sqrt{3} + 1 > \frac{44}{27}$.
$q < 0$ and $0\le r \le 1$: By Fact 1, we have $q^2 \ge 3r$. Thus, $q \le -\sqrt{3r}$. We have $|1 - q| + |1-r| = 1 - q + 1 - r = 2 - q - r \ge 2 + \sqrt{3r} - r \ge 2 > \frac{44}{27}$.
$q < 0$ and $r < 0$: We have $|1 - q| + |1-r| = 1 - q + 1 - r = 2 - q - r \ge 2 > \frac{44}{27}$.
Thus, the minimum is $\frac{44}{27}$.
$\phantom{2}$
Proof of Fact 1: We have \begin{align} q^2 &= (ab)^2 + (bc)^2 + (ca)^2 + 2abc(a+b+c)\\ &\ge ab\cdot bc + bc \cdot ca + ca \cdot ab + 2abc(a+b+c) \\ &= 3abc(a+b+c)\\ &= 3pr \end{align} where we have used \begin{align} &(ab)^2 + (bc)^2 + (ca)^2 - ab\cdot bc - bc \cdot ca - ca \cdot ab\\ =\ & \frac{1}{2}[(ab - bc)^2 + (bc - ca)^2 + (ca - ab)^2]\\ \ge \ & 0. \end{align}