Stuck with Lagrange Multiplier with no solution 
Find the minimum value of $f(x,y,z) = xy + 2yz + 2xz$ subject to the constraint $xyz = 108$.

I just tried normal lagrange multiplier and after some hard time I found only one solution which is $(6,6,3)$.
 With the value of $f(6,6,3) = 108$. But I think this doesn't prove anything.
It is not the max point neither the min point since I tried $(108,1,1)$ and $(-1,-1,108)$. What should I do?
 I used wolframalpha and it says that there is no min and max point.
Is this true? What should I do?
 Thank you in advance
 And sorry I do not know LaTex
Effort in the picture

Link Wolfram

 A: The thing is that $(6,6,3)$ is just a local minimum. All points close to this local minimum give a greater value of $f(x,y,z)$. However, this doesn't mean anything about the global behaviour of $f$.
To visualize this, lets look at a "slice" of $f(x,y,z)$. Consider
$$
\begin{aligned}
y &= x\\
z &= \frac{108}{xy}
\end{aligned}
$$
I chose $y=x$ since you obtained that the mimum complies that condition. However, instead of $z=x/2=y/2$ I only use the constraint $xyz=108$ in order to vary $x$ and look at what happens to $f$. This is, $f(x,y,z) = f\left(x,x,\frac{108}{x^2}\right)$ is a function of a single variable and looks like this:

The blue line is what happens for positive $x$. Clearly there is a local minimum, precisely at $x=6$ as you obtained. However, for negative $x$ (red line) $f$ has no lower bound. In fact, you can make $f$ as negative as you want if $x\to 0^-$. Hence, the "global minimum" does not exist.
The output of wolfram is consistent with this, no global minimum, but one local minimum.

Hope this helps!
