# 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).

• Note: I don't know Lagrange multipliers.
– user875449
Jan 23, 2021 at 9:39
• ah the ab,c being reals is a drawback , i have a proof when $min(a,b,c)\ge -1$ Jan 23, 2021 at 9:42
– user875449
Jan 23, 2021 at 9:45
• @AlbusDumbledore Yes it is. But it may help me and others, prove the general case.
– user875449
Jan 23, 2021 at 9:49
• Notice that the first part does not need absolute value, since $$\lvert 1-(ab+bc+ca)\rvert=\lvert (a+b+c)^2-(ab+bc+ca)\rvert=\lvert a^2+b^2+c^2+ab+bc+ca\rvert$$ And we have $$a^2+b^2+c^2+ab+bc+ca=\frac12\cdot \left[(a+b)^2+(b+c)^2+(c+a)^2\right]\geqslant 0$$ Jan 23, 2021 at 12:47

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:

1. $$q \ge 0$$: 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 $$a = b = c = \frac{1}{3}$$.

2. $$q < 0$$ and $$r > 1$$: By Fact 1, we have $$q^2 \ge 3$$. Thus, $$q < -\sqrt{3}$$. We have $$|1-q| + |1 - r| = 1 - q + r - 1 = -q + r > \sqrt{3} + 1 > \frac{44}{27}$$.

3. $$q < 0$$ and $$0\le r \le 1$$: By Fact 1, we have $$q^2 \ge 3r$$. Thus, $$q \le -\sqrt{3r}$$. We have $$|1 - q| + |1-r| = 1 - q + 1 - r = 2 - q - r \ge 2 + \sqrt{3r} - r \ge 2 > \frac{44}{27}$$.

4. $$q < 0$$ 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}

• Its a nice proof(+1) Jan 23, 2021 at 13:18
• This numerical treatment based on my partial answer gets $1.6294$, slightly less than $\tfrac{44}{27}\approx1.6296$. I take this to result from a rounding error, so see it as a sanity check supportive of your result.
– J.G.
Jan 23, 2021 at 13:19
• @AlbusDumbledore Thanks! Jan 23, 2021 at 13:41
• @RiverLi :) by the way if we put $x=ab,y=bc,z=ca$ then $q^2\ge 3pr$ is equivalent to $p^2\ge 3q$ so the Proof of fact 1 is not required Jan 23, 2021 at 13:43
• @AlbusDumbledore Nice! The first time I see that. Jan 23, 2021 at 13:48

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

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$$y\le y_0:=\frac{2+\sum_\pm\sqrt{2\pm\sqrt{3}}}{6}\approx0.7.$$ 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.