Show that $3\ln(x)+\frac{1}{x}$ has two zeros 
Show that $3\ln(x)+\frac{1}{x}$ has two zeros on $(0,\infty)$.


What came to my mind right away was to use the intermediate value theorem but the problem is that we now have to find two disjoint intervals $[a_1,b_1]$ and $[a_2,b_2]$ and prove that there is a zero in each of them, which hasn't worked out so far. I have also thought about integrating and applying Rolle's theorem to the antiderivative but this seems even more difficult.
Thank you very much in advance.
EDIT: The problem of find at least two zeros has been resolved due to the various answers and comments (thank you very much, everyone), however, proving that there are exactly two zeros seems to be the really difficult part here.
 A: Let $f(x)=3\ln x+\frac1x$. Evaluate 
$$f’(x)= \frac1x (3-\frac1x)$$
Note that $f’(x)<0$ for $x<\frac13$ and $f’(x)>0$ for $x>\frac13$. Therefore, $f(x)$ strictly decreases left of 1/3 and strictly increases right of 1/3. Also, we have $\lim_{x\rightarrow 0, \>\infty }f(x)=\infty$ and $f(1/3) <0$. 
Thus, $f(x)$ crosses 0 only once on either side of 1/3, hence only two roots.
A: Hint: Look at the intervals $[1/e,1]$ and $[1/e^2,1/e].$
A: Hint: $3 \ln(x) + 1/x$ has a minimum at $x = \ldots$, is increasing on $\ldots$ and decreasing on $\ldots$, and goes to $+\infty$ as $x \to \ldots$ and $\ldots$.
BTW, the two real solutions are $-\frac{1}{3 W_0(-1/3)}$ and $-\frac{1}{3 W_{-1}(-1/3)}$ where
$W_0$ and $W_{-1}$ are branches of the Lambert W function.
A: $\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
In terms of
well-known
Lambert $\operatorname{W}$ function,
which was introduced specifically to solve such equations,
\begin{align} 
3\ln(x)+\frac1{x}&=0
,\\
-3\ln(x)&=\frac1{x}
,\\
\ln\Big(\frac1x\Big)&=\frac1{3x}
,\\
\frac1x&=\exp\Big(\frac1{3x}\Big)
,\\
\frac1x\exp\Big(-\frac1{3x}\Big)
&=1
,\\
-\frac1{3x}\exp\Big(-\frac1{3x}\Big)
&=-\frac1{3}
,\\
\W\left(-\frac1{3x}\exp\Big(-\frac1{3x}\Big)\right)
&=\W\left(-\frac1{3}\right)
,\\
-\frac1{3x}
&=\W\left(-\frac1{3}\right)
,\\
x&=-\frac1{3\W\left(-\frac1{3}\right)}
.
\end{align} 
Since the argument of $\W\left(-\frac1{3}\right)$
is in a range $(-\frac1{\mathrm e},0)$,
there are two real solutions:
\begin{align} 
x_0&=-\frac1{3\Wp\left(-\frac1{3}\right)}
\approx 0.538449650261
,\\
x_1&=-\frac1{3\Wm\left(-\frac1{3}\right)}
\approx 0.220438937109
.
\end{align} 
$\endgroup$
A: A graphic solution. Since your functions are elementary, you may think about plotting them (by hand, yes. It's elementary).
You just want
$$3\ln(x) = -\frac{1}{x}$$
Hence you plot this two functions on the XY plane, obtaining:

Of course this only shows you have two real roots. If you want to know WHAT they are, then you will go with, guess, derivative or Newton's methods, Rolle theorem... whatever!
A: $x=e^y$, $y \in (-\infty, \infty)$.
$f(y)=3y +e^{-y}$;
$\lim_{y \rightarrow \pm \infty} f(y)=\infty$.
$f''(y)=e^{-y}>0$ , convex function has at most 2 zeroes.
$f'(y)=0$; $3=e^{-y}$; $y=-\log 3$;
$f(-\log 3)=-3\log 3+3<0$;
Exactly $2$ zeroes.
A: Hint. Show that it is anti-convex by seeing that the second derivative $$-\frac{3}{x^2}+\frac{2}{x^3}<0$$ for large $x.$ In particular this happens for $x>\frac23.$
Then show that there are no zeros in $(0,2/3].$
Next, show that the maximum occurs in $(2/3,+\infty),$ as opposed to, say, $[2/3,+\infty).$
Finally, examine the endpoints of $(2/3,+\infty).$
