Find the minimum and maximum value of $F=|a-2b|+|b-2c|+|c-2a|$

Question

Let $a, b, c \in \Bbb R$ satisfy $a^2+b^2+c^2=21$. Find the minimum and maximum value of $$F=|a-2b|+|b-2c|+|c-2a|$$

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I found $7\le F \le \sqrt{399}$ but couldn't prove it. I was thinking of the following inequality:

$$|x_1+x_2+\cdots+x_n|\le|x_1|+|x_2|+\cdots+|x_n|\le \sqrt{n \left(x_1^2+x_2^2+\cdots+x_n^2\right)}$$

but they are not really efficient. Does anyone know how to solve this problem or know where it first appeared?

Answer 1

The min and max bounds of $7$ and $\sqrt{399}$ are indeed correct. Let $$F_L = |a + b + c|$$ and $$F_U = \sqrt{15(a^2 + b^2 + c^2) - 12(ab + bc + ac)}$$ By the inequality you've proposed, we have $F_L \le F \le F_U$, and presumably you've performed the steps to show that $\min_{a^2 + b^2 + c^2 = 21} F_L = 7$ and $\max_{a^2 + b^2 + c^2 = 21}F_U = \sqrt{399}$; the values to obtain the min and max for these lower and upper bounds are $(a, b, c) = (4, 2, 1)$ and $(a, b, c) = (-3\sqrt{\frac{21}{19}}, \sqrt{\frac{21}{19}}, 3\sqrt{\frac{21}{19}})$. Substituting these in for $F$ also obtains the lower and upper bounds.

Answer 2

1) For the maximum

It is easy to prove that $$F = p a + qb + r c$$ for some $p, q, r$ (dependent on $a, b, c$) with $p^2 + q^2 + r^2 \le 19$. (See the remarks at the end.)

Using Cauchy-Bunyakovsky-Schwarz, we have $$pa + qb + rc \le \sqrt{(p^2 + q^2 + r^2)(a^2 + b^2 + c^2) } \le \sqrt{19 \cdot 21} = \sqrt{399}.$$

Also, when $a = \frac{3}{19}\sqrt{399}, b = -\frac{1}{19}\sqrt{399}, c = - \frac{3}{19}\sqrt{399}$, we have $a^2 + b^2 + c^2 = 21$ and $F = |a - 2b| + |b - 2c| + |c - 2a| = \sqrt{399}$.

Thus, the maximum of $F$ is $\sqrt{399}$.

Remarks:

We split into eight cases.

If $a - 2b \ge 0, b - 2c \ge 0, c - 2a \ge 0$, we have $$F = a - 2b + b - 2c + c - 2a = -a - b - c.$$

If $a - 2b \ge 0, b - 2c \ge 0, c - 2a < 0$, we have $$F = a - 2b + b - 2c - c + 2a = 3a - b - 3c.$$

If $a - 2b \ge 0, b - 2c < 0, c - 2a \ge 0$, we have $$F = a - 2b - b + 2c + c - 2a = -a - 3b + 3c.$$

Similarly, we deal with the remaining $5$ cases.

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$\phantom{2}$

2) For the minimum

Denote $x = a - 2b, y = b - 2c, z = c - 2a$.

We have \begin{align*} F^2 &= (|x| + |y| + |z|)^2\\ &= x^2 + y^2 + z^2 + 2|xy| + 2|yz| + 2|zx|\\ &\ge x^2 + y^2 + z^2 + 2|xy + yz + zx|\\ &= 105 - 4(ab + bc + ca) + |6(ab + bc + ca) - 84|. \end{align*} where we have used $x^2 + y^2 + z^2 = 5(a^2 + b^2 + c^2) - 4(ab + bc + ca)$, and $xy + yz + zx = 3(ab + bc + ca) - 2(a^2 + b^2 + c^2) $.

If $6(ab + bc + ca) - 84 \ge 0$, we have \begin{align*} F^2 &\ge 105 - 4(ab + bc + ca) + 6(ab + bc + ca) - 84\\ & = 21 + 2(ab + bc + ca)\\ &\ge 21 + 2 \cdot \frac{84}{6}\\ & = 49. \end{align*}

If $6(ab + bc + ca) - 84 < 0$, we have \begin{align*} F^2 &\ge 105 - 4(ab + bc + ca) - 6(ab + bc + ca) + 84\\ &= 189 - 10(ab + bc + ca) \\ &\ge 189 - 10\cdot \frac{84}{6} \\ &= 49. \end{align*}

Thus, we have $F \ge 7$.

Also, when $a = 4, b = 2, c = 1$, we have $a^2 + b^2 + c^2 = 21$ and $F = 7$.

Thus, the minimum of $F$ is $7$.

Answer 3

Maple: "extrema(F,g,{a,b,c},'s')" where $F:=|a-2b|+|b-2c|+|c-2a|$,

$g:=a^{2}+b^{2}+c^{2}=21$

‘s’ are the solutions.

$min=3\sqrt{7}$, $max=\sqrt{399}$.

The solutions are:

$a=\sqrt{7}, b=\sqrt{7}, c=\sqrt{7}$,

$a=-\sqrt{7}, b=-\sqrt{7}, c=-\sqrt{7}$;

$a=-\frac{3\sqrt{399}}{19}, b=+\frac{\sqrt{399}}{19}, c=+\frac{3\sqrt{399}} {19}$,

$a=-\frac{3\sqrt{399}}{19}, b=+\frac{3\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$,

$a=-\frac{\sqrt{399}}{19}, b=-\frac{3\sqrt{399}}{19}, c=+\frac{3\sqrt{399}}{19}$,

$a=+\frac{\sqrt{399}}{19}, b=+\frac{3\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$,

$a=+\frac{3\sqrt{399}}{19}, b=-\frac{3\sqrt{399}}{19}, c=+\frac{\sqrt{399}}{19}$,

$a=+\frac{3\sqrt{399}}{19}, b=-\frac{\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$.