MHB How Can You Maximize This Square Root Expression for All Real Numbers?

[anemone]

Here is this week's POTW:

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Maximize $\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}$ for all $x\in \Bbb{R}$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

No one answered last week's problem.(Sadface)

You can find the suggested solution as follows:

Spoiler

Let $f(x)=\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}=\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+x^2}$ and $P(x^2,\,x),\,A(2,\,3)$ and $B(1,\,0)$.

We are asked to maximize $\overline{PA}-\overline{PB}$. According to the triangle inequality, $\overline{PA}-\overline{PB}$ is at its maximum when the points $P$, $Q$ and $R$ lie on a straight line, therefore, we get

$\begin{align*}f(x)&=\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}\\&=\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+x^2}\\&=\overline{PA}-\overline{PB}\\&\le \overline{AB}\\&=\sqrt{(2-1)^2+(3-0)^2}\\&=\sqrt{10}\end{align*}$