<s>[INST] Largest value of the function $f(x) = \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$ Largest value of the function $f(x) = \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}\;\;,$ where $x\in \mathbb{R}$
My Try:: Let $y=f(x)=\sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$
$\displaystyle \Rightarrow y^2=(x^4-3x^2-6x+13)+(x^4-x^2+1)-2\sqrt{(x^4-3x^2-6x+13)\cdot (x^4-x^2+1)}$
$\displaystyle y^2=(2x^4-4x^2-6x+14)-2\sqrt{\{(x^2-2)^2+(x-3)^2\}\cdot\{(x^2-1)^2+\left(x\right)^2\}}$
Now using the Cauchy-Schwarz inequality for
$\displaystyle a=(x^2-2)^2+(x-3)^2$ and $\displaystyle b=(x^2-1)^2+\left(x\right)^2$
$\displaystyle \Rightarrow (a^2+b^2)\cdot(1^2+1^2)\geq(x+y)^2$ and equality hold when $\displaystyle\frac{a}{1}=\frac{b}{1}$
$\displaystyle\Rightarrow \left\{(x^2-2)^2+(x-3)^2\right\}\cdot \left\{(x^2-1)^2+\left(x\right)^2\right\}\geq \left\{(x^2-2)\cdot\left(x^2-1\right)+(x-3)\cdot x \right\}^2 = \left\{x^4-2x^2-3x+2\right\}$
and equality holds when $\displaystyle \frac{x^2-2}{x^2-1}=\frac{x-3}{x}\Rightarrow 3x^2-x-3=0$
So $\displaystyle \Rightarrow y^2\leq (2x^4-4x^2-6x+14)-2(x^4-2x^2-3x+2)=10$
So $y=f(x)\leq \sqrt{10}$ and equality holds when $\displaystyle x=\frac{1-\sqrt{37}}{6}$
If there is any geometrical method Then please explain here.
 [/INST] Hint
$$f(x)=\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$
let
$$A(x,x^2),B(3,2),C(0,1)$$
so
$$f(x)=|AB|-|AC|\le|BC|=\sqrt{10}$$

then we have
$$AB:y-1=\dfrac{1}{3}(x-0)$$
so
$$\begin{cases}
y=x^2\\
y=\dfrac{1}{3}x+1
\end{cases}$$
$$\Longrightarrow x=\dfrac{1-\sqrt{37}}{6}$$
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