Minimizing $|x_1-x_2|+|x_2-x_3|$ given two conditions 
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.
 A: 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 $\dfrac{4}{4+\sqrt{12}} = 4-\sqrt{12}.$
A: 
( vertical scale compressed )
A way to see this intuitively is to consider the parabolic curve $ \ y \ = \ x^2 \ \ . $  We are to choose three values of $ \ x \ $ so that the vertical spacings of their points on the curve are $ \ 2 \ $ units apart.  We see that since the parabola becomes "steeper" as $ \ x \ > \ 0 \ $ increases, the total spacing $ \ x_3 - x_1 \ $ decreases.  So we would want to have our three values of $ \ x \ $ be as large as possible in the interval, which means setting $ \ x_3 \ = \ 4 \ \ , $ for which we then must have $ \ x_2 \ =  \ \sqrt{14} \ $ and $ \ x_1 \ = \ \sqrt{12} \ \ . $
We can estimate the required spacing by using the slope formula $ \ m \ = \ \frac{\Delta y}{\Delta x} \ \ : \ $ at a particular value of $ \ x \ \ , $ we take the nearby points $ \ x \ \pm \ \delta \ $ to write
$$ m \ \ \approx \ \ \frac{(x + \delta)^2 \ - \  (x - \delta)^2}{(x + \delta) \ - \ (x - \delta)} \ \ = \ \ \frac{4·\delta·x}{2 · \delta} \ \ = \ \ 2·x \ \ . $$
The spacing between $ \ x_{i} \ $ and $ \ x_{i+1} \ $ will then be about
$ \ x_{i+1}  -  x_{i} \ = \ \frac{\Delta y}{m} \ = \  \frac{2}{2·x_{i}} \ = \ \frac{1}{x_{i}} \ \ , $ indicating that the spacing between $ \ x-$values will be "tighter" as $ \ x \ $ increases.  For purposes of comparison, with $ \ x_3 \ = \ 3 \ \ , $ we have roughly $ \ x_3 - x_1 \ = \ \frac{1}{\sqrt5} \ + \ \frac{1}{\sqrt7} \ \approx \ 0.825 \ \ , $ while for $ \ x_3 \ = \ 4 \ \ ,  \ x_3 - x_1 \ = \ \frac{1}{\sqrt{12}} \ + \ \frac{1}{\sqrt{14}} \ \approx \ 0.556 \ \ . $   This confirms our geometrical intuition.
Naturally, the precise answer is derived from the method used by dezdichado.
