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. 

For positive $x$, $y$, $z$, $a$, $b$, and $c$, note that that 
\begin{align*}
(x+y+z)^2 &= \left(\frac{x}{\sqrt{a}}\sqrt{a} + \frac{y}{\sqrt{b}}\sqrt{b} +\frac{z}{\sqrt{c}}\sqrt{c}\right)^2\\
&\le  \left(\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c}\right)(a+b+c).
\end{align*}
That is,
\begin{align*}
\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}.
\end{align*}
Then
\begin{align*}
\frac{a^3}{b} + \frac{b^3}{c} +\frac{c^3}{a} &=\frac{a^4}{ab} + \frac{b^4}{bc} +\frac{c^4}{ac}\\
&\ge \frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\\
&\ge \frac{\left(ab+bc+ac\right)^2}{ab+bc+ac}\\
&=ab+bc+ac.
\end{align*}
