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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)).

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  • $\begingroup$ @MrRipstein Hint: $x>0$ $\endgroup$ – Alexander Voliotis Apr 6 '18 at 18:07
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$\displaystyle f'(x)=e^\frac{a}{x}+xe^\frac{a}{x}\left(\frac{-a}{x^2}\right)=\frac{(x-a)e^\frac{a}{x}}{x}$.

$f'(x)>0$ 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 $g(x)=e^x-ex$.

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