TITLE: Find lower bound of function
QUESTION [5 upvotes]: Can someone help me finding a lower bound to the function
$$f(x)=\frac{x-1}{e^{-1}-xe^{-x^2}},$$
where $x\in[1,+\infty[$?
Taking the derivative and then solve $f'(x)=0$ isn't analytically possible. Then I tried the second best thing, find a lower bound but I don't really know how to start, so any help would be most welcome.

REPLY [2 votes]: Following njguliyev's comment, consider
$$ \frac{1}{f(x)}=\frac{g(x)-g(1) }{x-1}\tag{1}$$
where $g(x)=-xe^{-x^2}$. By the Mean value theorem, every value of $1/f$ on $(1,\infty)$ is also attained by $g'$ on $(1,\infty)$. So we need an upper bound on $g'$. Since
$$g'(x) = (2x^2-1)e^{-x^2}$$
$$g''(x) = 2x(3-2x^2)e^{-x^2}$$ 
it follows that $g'$ is maximal when $x^2=3/2$. Thus, the maximum of $g'$ on $(1,\infty)$ is $2e^{-3/2}$. This gives an upper bound on (1), and consequently 
$$f(x) \ge \frac{e^{3/2}}2 ,\quad x\ge 1$$
 This lower bound is $\approx 2.24$, not   far from the minimum of $\approx 2.34$ found in comments numerically.