I have to prove that $$f_a(x)=\ln x^2 +\frac{a}{x}$$ has no zeros if for $a>0$ it holds that $a\cdot e>2$.
I tried to find the zeros but I came to no result.
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1What did you try? Where do you get stuck? – TZakrevskiy Jan 29 '20 at 12:45
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$f$ is not defined at $0$. If $f$ is defined on $(0, \infty)$ we can argue as follows:
$f'(x)=\frac 2 x-\frac a {x^{2}}$ is positive for $x >\frac a 2$ and negative for $x <\frac a 2$. Hence $f$ has a minimum at $x =\frac a 2$. The minimum value is $2 \ln (\frac a 2)+2$ and this is positive if $ae >2$. Hence $f(x) >0$ for all $x$ if $ae >2$.
On $(-\infty, 0)$ the function $f$ vanishes at some point whatever $a$ is.
Kavi Rama Murthy
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In fact, the proposition is incorrect when allowing negative $x$, as $\lim_{x\to 0^-}f_a(x)=-\infty$ and $\lim_{x\to -\infty}f_a(x)=+\infty$ and $f_a(x)$ is continuous on $(-\infty,0)$. – Ingix Jan 29 '20 at 14:41
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