Given $f(x)=xe^{a \over x}, x>0$, for which it applies $f(x)\ge e^a, \forall x>0 $ (1)
• Show that $a=1.$
Personal work:
Even by getting $e^a$ to the other side of the equation it still, does not remind $xe^{a\over x}$. I'm baffled on how to find $a$ with the given relationship ((1)).
$\displaystyle f'(x)=e^\frac{a}{x}+xe^\frac{a}{x}\left(\frac{-a}{x^2}\right)=\frac{(x-a)e^\frac{a}{x}}{x}$.
$f'(x)>0$ for $x>a$ and $f'(x)<0$ for $0<x<a$.
$f$ attains its minimum at $x=a$.
So, $f(x)\ge f(a)=ae$ for $x>0$.
Therefore, $ae=e^a$.
Obviously, $a=1$ satisfies the relation.
It can be shown that $a=1$ is the only answer by considering the minimum value of $g(x)=e^x-ex$.