Minimum of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$ 
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).
 A: 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}
A: partial proof when  $a,b,c\ge -1$
WLOG $c=\min(a,b,c)\implies 3c-13<0$
$$|1-abc|+|1-ab-bc-ca|\ge |2-abc-ab-bc-ca|\ge 2-abc-ab-bc-ca$$ It remains to prove $$2-abc-ab-bc-ca\ge 44/27$$
$$\iff 2-ab(1+c)-c(a+b)-44/27\ge 0$$ $$2-\frac{{(a+b)}^2}{4}(1+c)-c(1-c)-44/27\ge 0$$
$$\iff 2-\frac{{(1-c)}^2}{4}(1+c)-c(1-c)-44/27\ge 0$$ $$\iff \frac{-1}{108}(3c-13){(3c-1)}^2\ge 0$$ which is obvious.
Update: River Li has a complete proof
A: I don't have a full solution either, but I thought I'd sketch out the theory regarding cubics this problem touches on.
The monic polynomial with roots $a,\,b,\,c$ is $t^3-t^2+yt-z$ with $y:=ab+bc+ca,\,z:=abc$. We want to minimise $\Sigma:=|1-y|+|1-z|$ subject to$$\Delta_3:=y^2-4y^3-4z-27z^2+18yz\ge0.$$This quadratic in $z$ can only be $\ge0$ if it has real roots $z_\pm:=\frac{9y-2\pm\sqrt{\Delta}}{27}$ with$$\Delta:=4(1-9y+36y^2-36y^3),$$whence $\Delta_3\ge0\iff z\in[z_-,\,z_+]$. You can show $\Delta$ is a cubic in $y$ which is $\ge0$ iff$$y\le y_0:=\frac{2+\sum_\pm\sqrt[3]{2\pm\sqrt{3}}}{6}\approx0.7.$$ The minimum of $\Sigma$ will occur at $\Delta_3=0$, i.e. $z=z_\pm(y)$.
Case I $z<1$ so $\Sigma=2-y-z$. Since $z_-\le\tfrac{9y-2}{27}<0.2$, for case I we minimise $2-y-z_+=\frac{56-36y-\sqrt{\Delta}}{27}$. This is an exercise in calculus.
Case II $z>1$ so $\Sigma=z_+-y=\tfrac{-18y-2+\sqrt{\Delta}}{27}$; again, there's a calculus exercise.
Both of the above calculus exercises require us to solve cubics.
