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?

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.
