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. 

First I'll prove a Lemma: $a^2+b^2+c^2\ge ab+bc+ca$ for all $a,b,c\in\mathbb R$.
Proof: it follows from the Rearrangement Inequality, because $(a,b,c)$ and $(a,b,c)$ are similarly sorted.
Or notice that it's equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true; or add the following inequalities: $a^2+b^2\ge 2ab$, $b^2+c^2\ge 2bc$, $c^2+a^2\ge 2ca$. $\ \square$
Your inequality is cyclic. Wlog there are two cases:

*

*$a\ge b\ge c$. Then $(a,b,c)$ and $(1/a,1/b,1/c)$ are oppositely sorted, so
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge \frac{a^3}{a}+\frac{b^3}{b}+\frac{c^3}{c}=a^2+b^2+c^2$$
Now use the Lemma.


*$a\ge c\ge b$. Then $(a,c,b)$ and $(1/a,1/c,1/b)$ are oppositely sorted, so  $$\frac{a^3}{b}+\frac{c^3}{a}+\frac{b^3}{c}\ge \frac{a^3}{a}+\frac{c^3}{c}+\frac{b^3}{b}=a^2+c^2+b^2$$
Now use the Lemma.
In your inequality, equality holds if and only if $a=b=c$.

Another proof: By Hölder's inequality:
$$\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge \frac{(a+b+c)^3}{3(a+b+c)}=\frac{(a+b+c)^2}{3}$$
Also $(a+b+c)^2\ge 3(ab+bc+ca)$, because this is equivalent to $a^2+b^2+c^2\ge ab+bc+ca$ (see the Lemma).
