Proving that $\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}> \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ 
Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$  are different positive real numbers.

First, I tried to simplify the proof statement but I got an even more complicated: $$a^4b^4+b^4c^4+a^4c^4> a^2b^3c^3+b^2c^3a^3+a^3b^3c^2$$
Then I used Power mean inequality on $\dfrac{1}{a},\dfrac{1}{b},\dfrac {1}{c}$ but that wasn't useful here.
Finally, I tried to solve it using AM-HM inequality but couldn't.
What would be an efficient method to solve this problem? Please provide only a hint and not the entire solution since I wish to solve it myself.
 A: AM-GM helps!
$$\sum_{cyc}\frac{ab}{c^3}=\frac{1}{4}\sum_{cyc}\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)\geq\frac{1}{4}\sum_{cyc}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}\right)=\sum_{cyc}\frac{1}{c}.$$
Done!
Without $cyc$ we can write the solution so:
$$\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}=$$
$$=\frac{1}{4}\left(\left(\frac{2ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{b^3}\right)+\left(\frac{ab}{c^3}+\frac{2bc}{a^3}+\frac{ca}{b^3}\right)+\left(\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{2ca}{b^3}\right)\right)\geq$$
$$\geq\frac{1}{4}\left(4\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}+4\sqrt[4]{\left(\frac{bc}{a^3}\right)^2\cdot\frac{ab}{c^3}\cdot\frac{ca}{b^3}}+4\sqrt[4]{\left(\frac{ca}{b^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ab}{c^3}}\right)=$$
$$=\frac{1}{c}+\frac{1}{a}+\frac{1}{b}.$$
The same trick gives also a proof by Holder:
$$\sum_{cyc}\frac{ab}{c^3}=\sqrt[4]{\left(\sum_{cyc}\frac{ab}{c^3}\right)^2\sum_{cyc}\frac{bc}{a^3}\sum_{cyc}\frac{ca}{b^3}}\geq\sum_{cyc}\sqrt[4]{\left(\frac{ab}{c^3}\right)^2\cdot\frac{bc}{a^3}\cdot\frac{ca}{b^3}}=\sum_{cyc}\frac{1}{c}.$$
Turned out even a bit of shorter.
A: Use the rearrangement inequality. Assume without loss of generality $a\geq b\geq c>0$. Then we have $ab\geq ac\geq bc\,$ and $\,1/c^3\geq 1/b^3\geq 1/a^3$. Therefore the sorted sum-product
$$\frac{ab}{c^3}+\frac{ac}{b^3}+\frac{bc}{a^3}\geq\frac{bc}{c^3}+\frac{ab}{b^3}+\frac{ac}{a^3}=\frac{b}{c^2}+\frac{a}{b^2}+\frac{c}{a^2}$$
is greater than equal to a shuffled sum-product, which is greater than equal to the reversed sum-product
$$\frac{b}{c^2}+\frac{a}{b^2}+\frac{c}{a^2}\geq\frac{c}{c^2}+\frac{b}{b^2}+\frac{a}{a^2}=\frac{1}{c}+\frac{1}{b}+\frac{1}{a}.$$
Equality is obtained if and only if $\,a=b=c$. If $a,b,c$ are different (not necessarily all different), the greater than "$>$" sign holds.
A: \begin{eqnarray*}
a^4(b^2-c^2)^2+b^4(a^2-c^2)^2+c^4(b^2-a^2)^2+a^2b^2c^2((a-b)^2+(c-b)^2+(a-c)^2) \geq 0.
\end{eqnarray*}
Rearrange to 
\begin{eqnarray*}
a^4b^4+b^4c^4+c^4a^4 \geq a^2b^2c^2(ab+bc+ca).
\end{eqnarray*}
Now divide by $a^3b^3c^3$ and we have
\begin{eqnarray*}
\frac{ab}{c^3}+\frac{bc}{a^3}+\frac{ca}{ b^3}  \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.
\end{eqnarray*}
