$f(x)=\sqrt{(x-1)^2+(x^2-5)^2}\;\;,\;\; g(x)=\sqrt{(x+2)^2+(x^2+1)^2},\forall x\in \mathbb{R}$, Then Max $\left\{f(x)-g(x)\right\}$ Let $f(x)=\sqrt{(x-1)^2+(x^2-5)^2}\;\;,\;\; g(x)=\sqrt{(x+2)^2+(x^2+1)^2},\forall x\in \mathbb{R}$.
Find the Minimum of function $\left\{f(x)+g(x)\right\}$ and the maximum of function $\left\{f(x)-g(x)\right\}$. 
$\bf{My\; Try}$:: For Minimum of  $\left\{f(x)+g(x)\right\}$
Using Minkowski inequality or $\triangle$ Inequality $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\geq \sqrt{(a+b)^2+(c+d)^2}$
and equality holds when $\displaystyle \frac{a}{b} = \frac{c}{d}$
$\sqrt{(1-x)^2+(5-x^2)^2} + \sqrt{(x+2)^2+(x^2+1)^2}$
$\geq \sqrt{\left(1-x+x+2\right)^2+\left(5-x^2+x^2+1\right)^2} = \sqrt{3^2+6^2} = 3\sqrt{5}$
and equality hold, when $\displaystyle \frac{1-x}{5-x^2} = \frac{x+2}{x^2+1}$
But I did not understand How can I calculate Maximum of $\left\{f(x)-g(x)\right\}$
Help Required
Thanks

Use the same inequality again, but backwards:
$$\sqrt{(x+2)^2 + (x^2+1)^2} + \sqrt{((x-1)-(x+2))^2 + ((x^2-5)-(x^2+1))^2} \geq \sqrt{(x-1)^2 + (x^2-5)^2}$$
This gives us $f(x)-g(x) \leq 3\sqrt{5}$.  It also allows for us to solve for $x$.
One warning: this method is really finding the maximum value of $|f(x)-g(x)|$.  So when we solve for $x$, we will get two solutions, one maximizing $f(x)-g(x)$, the other maximizing $g(x)-f(x)$.
