calculusfunctionsoptimization
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.
Have a look at following picture. Does this help? Can you get (using this picture) area of the rectangle?
You should get $A(x)=2x(12-x^2)$ as mentioned in André's comment.
After getting this expression for $A(x)$, can you continue with the solution? What will be $A'(x)$ and where is it equal to zero?
You have $A(x)=24x-2x^3$, which means $$A'(x)=24-6x^2.$$ Solving $A'(x)=0$ gives $x=2$ and $A(x)=2\cdot2\cdot(12-2^2)=2\cdot2\cdot8=32$ as you correctly stated in the comment.
(There is also solution $x=-2$ but that one is not interesting for us - we are looking for $x\ge 0$ since we denote by $x$ the point to the right from $0$.)
As indicated in my comment, you can show that $g(x)=x-\log x$ is a lower-bound function to $f(x)=x/x^{1/x}$.
For a proof, observe that we can write $g(x)=x\left(1-\log x/x\right)$ and $f(x)=x\exp\left(-\log x/x\right)$. Since $\exp y>1+y$ for all real $y$ it follows, by setting $y=-\log x/x$, that $\exp\left(-\log x/x\right)$ is greater than $1-\log x/x$ and thus $f(x)>g(x)$ for $x>0$.
In fact, we can also show that the difference between $f(x)$ and $g(x)$ approaches zero as $x$ tends to infinity. This follows from the expansion $\exp y=1+y+\frac12y^2+\mathcal{O}_{y\to 0}(y^3)$ which implies that $$ \exp\left(-\log x/x\right) - \left(1-\log x/x\right) = \frac12\left(\frac{\log(x)}{x}\right)^2 + \mathcal{O}_{x\to\infty}\left(\left(\frac{\log(x)}{x}\right)^3 \right). $$ In particular, $f(x)-g(x)\sim_{x\to\infty}\frac12\log x^2/x$.
Numerically, the maximum of $f(x)-g(x)$ for $x>1$ occurs at approximately $x=8.1$ with a value of approximately $0.25$, so the approximation is indeed quite good.
Best Answer
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.