<s>[INST] 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.
 [/INST] 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.
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