TITLE: Find the maximum value of $ \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1} $
QUESTION [11 upvotes]: If $x\in\mathbb{R}$  find the maximum value of   
$$ \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1} $$  

I tried this:
  Let $$y= \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$$
  For maxima $\frac{dy}{dx}=0$ and $\frac{d^2y}{dx^2} < 0$. However, the equation $\frac{dy}{dx}=0$ (after simplifying and clearing the square roots) came out to be a nine degree equation which gave me a nightmare! Moreover, simplifying the derivative was also a tedious task. I found this question in my book in the chapter on theory of equation. I can't think of an algebraic solution. Please Help!
  Thanks!

REPLY [15 votes]: since
$$\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$
let $$P(x,x^2),A(3,2),B(0,1)$$
so
$$|PA|-|PB|\le |AB|=\sqrt{10}$$
if and only is $A,P,B$ on a line.