Prove that the function $(12-6x+x^2) e^x - (12+6x+x^2)$ is positive on $(0, \infty)$ 
Prove that the function $(12-6x+x^2) e^x - (12+6x+x^2)$ is positive on $(0, \infty)$. 

I am unable to prove this.
 A: Let $f(x)=x+\ln(x^2-6x+12)-\ln(x^2+6x+12)$.
Since $f'(x)=1+\frac{2x-6}{x^2-6x+12}-\frac{2x+6}{x^2+6x+12}=\frac{x^4}{(x^2-6x+12)(x^2+6x+12)}\geq0$, 
we obtain that $f(x)\geq f(0)=0$.
Done!
A: We may notice that
$$ (12-6x+x^2)e^x-(12+6x+x^2)=\int_{0}^{x}\left[(z^2-4z+6)e^z-(6+2z)\right]\,dz $$
hence it is enough to prove that $(z^2-4z+6)e^z-(6+2z)$ is a positive function on $\mathbb{R}^+$.
In a similar way,
$$ (z^2-4z+6)e^z-(6+2z) = \int_{0}^{z}\left[(t^2-2t+2)e^t-2\right]\,dt $$
hence it is enough to prove that
$$ \left(1-t+\frac{t^2}{2}\right) e^t > 1 $$
for any $t>0$. But over such a set $1-t+\frac{t^2}{2}>e^{-t}$, hence the last inequality is trivial.

An alternative approach: you may check that all the coefficients of the Taylor series of $f(x)=(12-6x+x^2)e^x-(12+6x+x^2)$ centered at $x=0$ are non-negative, due to
$$ \frac{12}{(n+2)!}-\frac{6}{(n+1)!}+\frac{1}{n!}=\frac{(n-1)(n-2)}{(n+2)!}$$
hence for any $x>0$
$$ f(x) = \frac{x^5}{60}+\frac{x^6}{120}+\frac{x^7}{420}+\frac{x^8}{2016}+\ldots > 0.$$

Summarizing, we actually recognized in $\frac{12+6x+x^2}{12-6x+x^2}$ a Padé approximant for $e^x$. We may prove the original inequality also by noticing that
$$ \int_{0}^{t}x^2(t-x)^2 e^{x}\,dx \geq 0 $$
and computing the LHS.
A: HINT: the function is continuous because is a sum of two continuous functions. 
Then notice that exist a positive value (by example $f(1)>0$), if exist any negative value then by the intermediate value theorem must exist some $x_0$ such that $f(x_0)=0$. You can show that this is false for the given domain so the function is positive, i.e., doesnt exist any negative value.
