Find the minimum of $P = (a - b)(b - c)(c - a)$

Question

Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of $$P = (a - b)(b - c)(c - a)$$

My solution:

• We have:

$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$ $$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$ $$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$

• Using AM-GM Inequality, we have:

$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$ $$\implies 3 \sqrt[3]{P^2} \leq 12$$ $$\implies -8 \leq P \leq 8$$

• Therefore, $\min P = -8$

Is this solution correct? If not, then why?

Answer 1

WLOG $a\ge b\ge c$ and let $x=a-b,y=b-c,z=c-a$

We observe $x+y+z=0$ with $x,y\ge 0$.and as you have found $x^2+y^2+z^2=12$

Elimination of $z$ results in: ${(x+y)}^2=6+xy ...(1)$

since ${(x+y)}^2\ge 4xy$ which means $0\le xy\le 2$

let $xy=t$

since $0\le t\le 2$

Now $x^2y^2z^2=t^2(6+t)\le 6.2^2+2^3=32$....(using (1) and $z=-(x+y)$)

or $|xyz|\le 4\sqrt{2}$

Answer 2

The inequality $$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$ become equality when $a-b=b-c=c-a,$ or $a=b=c,$ but for this value then $$(a-b)(b-c)(c-a)=0 \ne -8.$$ This is my solution, we have $$P^2 = \frac{4(a^2+b^2+c^2-ab-bc-ca)^3-(a+b-2c)^2(b+c-2a)^2(c+a-2b)^2}{27} \quad (1)$$ $$\leqslant \frac{4}{27}(a^2+b^2+c^2-ab-bc-ca)^3.$$ Therefore $$P^2 \leqslant \frac{4}{27}(a^2+b^2+c^2-ab-bc-ca)^3 = 32,$$ or $$-4\sqrt 2 \leqslant (a-b)(b-c)(c-a) \leqslant 4\sqrt 2.$$ So $P_{\min} = -4\sqrt 2,$ equality occur when $a=1,\;b=1+2\sqrt 2,\;c=1+\sqrt 2.$

Note. How to find constant $\frac{4}{27}?$

For $(a-b)(b-c)(c-a) \ne 0,$ setting $x=a-b,\,y=b-c,$ then $$F = \frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2+c^2-ab-bc-ca)^3} = \frac{x^2y^2(x+y)^2}{(x^2+xy+y^2)^3}.$$ From $x^2+xy+y^2 \geqslant \frac{3}{4}(x+y)^2$ and the AM-GM inequality, we have $$F \leqslant \frac{64}{27} \cdot \frac{x^2y^2}{(x+y)^4} \leqslant \frac{4}{27}.$$ From this proof we get $$\frac{4}{27}- \frac{x^2y^2(x+y)^2}{(x^2+xy+y^2)^3}=\frac{(x-y)^2(2y+x)^2(2x+y)^2}{(x^2+xy+y^2)^3}.$$ It's equivalent to the identity $(1).$

Answer 3

We need to show that the value $-8$ occurs, otherwise, we can not say that it's a minimal value.

By the way, we can prove that the minimum is $-4\sqrt2.$

Indeed, we need to prove that $$(a-b)(b-c)(c-a)\geq-4\sqrt2\left(\sqrt{\frac{\sum\limits_{cyc}(a^2-ab)}{6}}\right)^3,$$ which easy to get by AM-GM after substitution $a-b=x$, $b-c=y$.

Answer 4

Using $x=a-b,y=b-c$, then $-(x+y)=c-a$, so $P=-xy(x+y)=-(x^2y+xy^2)$.

The condition $(a−b)^2+(b−c)^2+(c−a)^2=12$ becomes $g=x^2+y^2+xy=6$.

Now using Lagrange multipliers: \begin{align} \nabla P&=-\langle2xy+y^2,x^2+2xy\rangle\\ &=-\langle y(2x+y),x(x+2y)\rangle\\ \\ \nabla g&=\langle2x+y,x+2y\rangle \end{align}

Now letting $$\nabla P=\lambda\cdot\nabla g$$ $$\lambda=-y,\,\lambda=-x$$ So $x=y$, plugging this back in $g$: $$3x^2=6$$ $$x=\pm\sqrt2$$ Then the extreme values of $P$ are: $$P=-2\cdot\pm2\sqrt2=\pm4\sqrt2$$

Answer 5

Minimize $P = (a-b)(b-c)(c-a)$

given $(a - b)^2 + (b - c)^2 + (c - a)^2 = 12$

WLOG, say $a \ge b \ge c, \, a - b = x, a - c = y$ where $x \ge 0, y \ge x$

which gives,

$P (x,y) = xy(x-y)$ ...(i)

$\, G(x,y) = x^2 + y^2 + (x-y)^2 - 12 = 0$ ...(ii)

Using Lagrange,

$Q(x,y) = P (x,y) + \lambda G(x,y) = xy(x-y) + \lambda (2x^2 + 2y^2 -2xy - 12)$

You get $\, 2xy-y^2 + \lambda(4x - 2y) = 0$ ..(iii)

$\, x^2-2xy + \lambda(4y - 2x) = 0$ ...(iv)

From (iii) and (iv),

You get $y = 2x, x = 2y$

Taking $y = 2x$ (as $y \ge x$) and substituting in (ii), you get $x = \sqrt 2$

From (i) that gives min $\, P = -4\sqrt2$

Answer 6

Let $x = a - b, y = b-c, z = c-a$. Then we have $x^2 + y^2 + z^2 = 12$ and $x + y + z = 0$. Also, we have $xy + yz + zx = \frac{(x+y+z)^2 - (x^2+y^2+z^2)}{2} = -6$. We need to find the minimum of $xyz$.

Let $p = x + y + z = 0, q = xy + yz + zx = -6$ and $r = xyz$. It is easy to prove that $$-4p^3r+p^2q^2+18pqr-4q^3-27r^2 = (x-y)^2(y-z)^2(z-x)^2 = 27(32-r^2)\ge 0$$ which results in $-4\sqrt{2} \le r \le 4\sqrt{2}$. Also, when $x = \sqrt{2}, y = -2\sqrt{2}, z= \sqrt{2}$ (e.g. $a = -\sqrt{2}, b= -2\sqrt{2}, c = 0$), we have $xyz = -4\sqrt{2}$. Thus, the minimum of $xyz$ is $-4\sqrt{2}$.

Remark: Actually, $-4p^3r+p^2q^2+18pqr-4q^3-27r^2$ is the discriminant of $u^3 - pu^2 + qu - r = 0$.