Let $x_1, x_2, x_3 \in \Bbb R$, satisfy $0 \leq x_1 \leq x_2 \leq x_3 \leq 4$. If their squares form an arithmetic progression with common difference $2$, determine the minimum possible value of $$|x_1-x_2|+|x_2-x_3|$$
So far, I've started with the fact $x_2^2 – x_1^2 = x_3^2 – x_2^2 = 2$ since we know their squares form an arithmetic progression with common difference $2.$ We can solve this to obtain
$$x_2 = \pm\sqrt{x_1^2+2}, \qquad x_3 = \pm\sqrt{x_1^2+4}$$
I'm not sure how to continue. Thanks in advance for the help.
Best Answer
Notice that:
$$x_3 - x_1 = \dfrac{x_3^2-x_1^2}{x_3+x_1} = \dfrac{4}{\sqrt{x_1^2+4}+x_1}$$ and this is obviously minimized at the largest possible value of $x_1.$ That value is obtained by observing: $$16\geq x_3^2 = x_1^2+4\iff x_1\leq\sqrt{12}$$ and so the minimum is $\dfrac{4}{4+\sqrt{12}} = 4-\sqrt{12}.$