Skunky Maxima

$\begin{aligned} S & = \bigg| \sqrt{x^2 +4 x+5} - \sqrt{x^2+2 x+5} \bigg| \\ & = \bigg| \sqrt{(x +2)^2+1} - \sqrt{(x+1)^2+4} \bigg| \\ & = \bigg| \sqrt{(x - (-2))^2+(0 \pm1)^2} - \sqrt{(x-(-1))^2+(0 \pm 2)^2} \bigg| \end{aligned}$

Let the points be $P(x,0)$ on the x-axis, $A(-2,\pm 1)$ and $B(-1,\pm 2)$ .

Then $\sqrt{(x - (-2))^2+(0 \pm1)^2} = [AP]$ , the length of $AP$ , $\sqrt{(x-(-1))^2+(0 \pm 2)^2} = [BP]$ , the length of $BP$ and $S = |[AP]-[BP]|$ , the differences of the two lengths.

$S$ is maximum, when $A$ , $B$ and $P$ is in a straight line.

Therefore, $S_{max} = \sqrt{(-2+1)^2+(1-2)^2} = \sqrt{2}$ and $S_{max}^4 = \boxed{4}$ .

$S = |~\sqrt{x^2+4x+5} - \sqrt{x^2+2x+5}~|=|~\sqrt{(x+2)^2+(0 \pm 1)^2} - \sqrt{(x+1)^2+(0 \pm 2)^2}~| \\ If~~M(-2,\pm 1),~~~N( - 1, \pm ~ 2),~~~Q(x,0)~~~are~~three~point~in~an~x-y~ plain,\\ we~ see~that~S~represents~distance~ |~QM~-~QN~|. \\ This~distance~ would~ be~minimum~if~all~three~are~in~a~straight~line.\\ \implies~slope~of~QM = slope~of~QN.~~~\implies~\dfrac{\pm 1 - 0}{ - 2 - x}=\dfrac{\pm 2 - 0}{ -1 - x}.\qquad \qquad \therefore~x_{min}= - 3.\\ Substituting~x~in~S,~~~S_{min}=|\sqrt{9-12+5} - \sqrt{9-6+5}|=|\sqrt2-2*\sqrt2|=|-\sqrt2|=\sqrt2.\\ So~\Large S_{min}^4=4.$