If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.

Note that 
$$\frac{f(x)}{g(x)}=\frac{(x-1/x)^2+2}{x-1/x}=x-\frac1x+2\frac{1}{x-\frac1x}.
$$
Letting $u=x-1/x$ we take the derivative of $u+2/u$ to get $u'-\frac{2}{u^2}u'=u'(1-2/u^2)=0$.  As $u'$ is always positive we must have $2=u^2=(x-1/x)^2$ which is easily solved; $x=\pm\sqrt{2\pm\sqrt3}$.
