If $0Given that $f(x)=e^x(x^2-6x+12)-(x^2+6x+12),\;\;x>0$ is an increasing function. I want to prove that:


If $0<x<\infty$, then $0<\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}<\frac{x}{12}$. 


Here is what I have done:
If $0<x<\infty$, then by Mean Value Theorem, $\exists\; c\in(0,x)$ such that
$$f'(c)=\frac{e^x(x^2-6x+12)-(x^2+6x+12)- 0}{x- 0}>0$$
but how do I get the desired inequality? Can anyone help out?

We'll prove that $$0<\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}$$ or
$$\frac{1}{e^x-1}>\frac{2-x}{2x},$$ which is obvious for $x\geq2$.
But, for $0<x<2$ we need to prove that
$$e^x-1<\frac{2x}{2-x}$$ or $f(x)>0,$ where
$$f(x)=\ln(x+2)-\ln(2-x)-x.$$
Indeed, $$f'(x)=\frac{x^2}{4-x^2}>0,$$ which says
$$f(x)>\lim_{x\rightarrow0^+}f(x)=0$$ and the left inequality is proven.
By the same way we can prove a right inequality.
Indeed, we need to prove that
$$\frac{1}{e^x-1}<\frac{x}{12}+\frac{1}{x}-\frac{1}{2}$$ or 
$$e^x-1>\frac{12x}{x^2-6x+12}$$ or $g(x)>0,$ where
$$g(x)=x-\ln(x^2+6x+12)+\ln(x^2-6x+12)$$ and since
$$g'(x)=\frac{x^4}{(x^2+6x+12)(x^2-6x+12)}>0,$$ we obtain:
$$g(x)>\lim_{x\rightarrow0^+}g(x)=0$$ and we are done.
