Find the biggest value of $|a-b|+|b-c|+|c-a|$ when $a^2+b^2+c^2=1$

Question

Given that $a^2 + b^2 + c^2 = 1$ How can I prove that the maximum value of |a - b| + |b - c| + |c - a| is $2\sqrt{2}$ ? In extends, given that $a_1^2 + a_2^2 + a_3^2 + ... + a_{2020}^2 = 1$ find the maximum value of $|a_1 - a_2| + |a_2 - a_3| +|a_3 - a_4| + ... + |a_{2020} - a_1|$

Answer 1

Without loss of generality, assume that $a\geq b\geq c$. Then:

$$|a-b|+|b-c|+|c-a| = 2(a-c)$$

And we have:

$$ \begin{aligned} (a-c)^2&=2(a^2+c^2)-(a+c)^2\\ &\leq 2(a^2+c^2) \\ &\leq 2(a^2+b^2+c^2)\\ &=2 \end{aligned}$$

Therefore $2(a-c)\leq 2\sqrt{2}$. Equality occurs when $(a,b,c) = \left(\frac{1}{\sqrt{2}}, 0,-\frac{1}{\sqrt{2}}\right)$.