Q:

If $f,g:\Bbb R\to\Bbb R$ are distinct linear functions which map $[-1,1]$ onto $[0,2]$ and $h:\Bbb R\setminus\{-1,0,1\}\to\Bbb R$ defined by $h=f/g$

Given  two distinct linear functions $f$ and $g$ defined on $\mathbb R$ such that they map $[-1,1]$ onto $[0,2]$ and $h:\mathbb R\backslash \{-1,0,1\} \to \mathbb R$ defined by $h(x)=\frac{f(x)}{g(x)}$, show that 
$$
\left\lvert \,h\big(h\left(x\right)\big)+h\left(h\left(\frac{1}{x}\right)\right)\,\right\rvert > 2
$$
Attempts: 
Linear are the functions of the form $ax+by+c=0$.
Between $[-1,1]$ to $[0,2]$ since they are onto they must be a strictly increasing/decreasing.
I was not able to make an conclusions, also tried graphical approach.

A:

To proof your inequality :
First determinate $f$ and $g$ for that, you know that either $f$ will be increasing and $g$ will be decreasing or $g$ will be increasing and $f$ will be decreasing, so we can suppose that $f$ is increasing and $g$ is decreasing, so 
we have the system :
$$
\left\{\begin{array}{l} f(-1)=-a+b=0 \\
f(1)=a+b=2 \\
g(-1)=-c+d=2 \\
g(1)=c+d=0\\ \end{array}\right.
$$
Where $f(x)=ax+b$ and $g(x)=cx+d$
So $f(x)=x+1$ and $g(x)=-x+1$ so finally :
$$
h(x)=\frac{x+1}{-x+1}
$$
so $h(h(x))= \frac{-1}{x}$ and $h(h(\frac{1}{x}))=-x$.
we need now to proof that for all $a\in\mathbb{R}-\{0,1,-1\}$ 
$$
\left|a+\frac{1}{a}\right|> 2 
$$
in fact 
$$
\left(a+\frac{1}{a}\right)^2=a^2+2+\frac{1}{a^2}> 2+2=4
$$
because $(a-\frac{1}{a})^2> 0 \implies a^2+\frac{1}{a^2}> 2$
So 
$$
\left|h(h(x))+h(h(1/x))\right|=\left|x+\frac{1}{x}\right|> 2
$$
If now we suppose now that $g$ will be increasing and $f$ will be decreasing so we find $h'(x)=h(\frac{1}{x})$ and all calculation will be true.

