Prove that $\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c$ How to prove that  
\begin{equation*}\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c,\  where \ a,b,c>0\end{equation*}
I tried the following:
\begin{equation*}abc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge a+b+c\end{equation*}
Using Chebyshev's inequality
\begin{equation*}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\le3(\frac{1}{a}\frac{1}{a}+\frac{1}{b}\frac{1}{b}+\frac{1}{c}\frac{1}{c})\end{equation*}
from first inequality follows
\begin{equation*}\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2abc\ge a+b+c\end{equation*}
equivalent to
\begin{equation*}\frac{abc}{3(a+b+c)}\ge (\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}
and by amplifying  both members by 9
\begin{equation*}\frac{3abc}{a+b+c}\ge (\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}
now using mean inequality
\begin{equation*}\sqrt[3]{abc}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\end{equation*}
the inequality in question becomes
\begin{equation*}3abc\ge (a+b+c)(\sqrt[3]{abc})^2\end{equation*}
which yields
\begin{equation*}3\sqrt[3]{abc}\ge a+b+c\end{equation*}
not what I wanted.
 A: Multiplying both sides of your inequality with $abc>0$, you get equivalently that:
$ \displaystyle (ab)^2 + (bc)^2 + (ca)^2 \geq abc (a+b+c) $ 
This holds by the basic inequality $ \displaystyle x^2 + y^2 + z^2 \geq xy +yz+ xz $, which holds for all $x,y,z $ real.
edit: The basic inequality holds for all real, thank's to user26486, for pointing this out.
A: The inequality which you want to prove is symmetric, so we can assume without loss of generality that $a\geq b\geq c$.
Then $ab\geq ac\geq bc$ and $\frac{1}{c}\geq \frac{1}{b}\geq \frac{1}{a}$.
Thus, from the rearrangement inequality, we get that
$$\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq\frac{ab}{b}+\frac{ac}{a}+\frac{bc}{c}=a+c+b$$
which is what we wanted to show.
A: We can do a slick AM-GM "pairwise" token that I picked up from the "Cauchy masters":
$\dfrac{ab}{c} + \dfrac{bc}{a} \geq 2\sqrt{\dfrac{ab}{c}\cdot\dfrac{bc}{a}}= 2b$, and similarly: $\dfrac{bc}{a}+\dfrac{ca}{b} \geq 2c$, and $\dfrac{ca}{b} + \dfrac{ab}{c} \geq 2a$. Add them up and divide by $2$ to get the answer.
