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


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?
 [/INST] 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:
$a=\sqrt{7}, b=\sqrt{7}, c=\sqrt{7}$,
$a=-\sqrt{7}, b=-\sqrt{7}, c=-\sqrt{7}$;
$a=-\frac{3\sqrt{399}}{19}, b=+\frac{\sqrt{399}}{19}, c=+\frac{3\sqrt{399}}
{19}$,
$a=-\frac{3\sqrt{399}}{19}, b=+\frac{3\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$,
$a=-\frac{\sqrt{399}}{19}, b=-\frac{3\sqrt{399}}{19}, c=+\frac{3\sqrt{399}}{19}$,
$a=+\frac{\sqrt{399}}{19}, b=+\frac{3\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$,
$a=+\frac{3\sqrt{399}}{19}, b=-\frac{3\sqrt{399}}{19}, c=+\frac{\sqrt{399}}{19}$,
$a=+\frac{3\sqrt{399}}{19}, b=-\frac{\sqrt{399}}{19}, c=-\frac{3\sqrt{399}}{19}$.
</s>