<s>[INST] Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$. 
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.

I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work here. The $ab+bc+ca$ reminds of a cyclic expression, so that may help by factoring the inequality and getting a true statement. 
 [/INST] $$
\begin{align}
&\left(\frac{a^3}b+\frac{b^3}c+\frac{c^3}a\right)-(ab+bc+ca)\\
&=\frac ab\left(a^2-b^2\right)+\frac bc\left(b^2-c^2\right)+\frac ca\left(c^2-a^2\right)\\[3pt]
&=\left(\frac ab-1\right)\left(a^2-b^2\right)+\left(\frac bc-1\right)\left(b^2-c^2\right)+\left(\frac ca-1\right)\left(c^2-a^2\right)\\
&=\frac{(a-b)^2(a+b)}b+\frac{(b-c)^2(b+c)}c+\frac{(c-a)^2(c+a)}a\\[6pt]
&\ge0
\end{align}
$$
This also shows that equality requires $a=b=c$.
</s>