Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$

The next thing you can try is Cauchy again $$\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)(b+c+a)\geq (a+b+c)^2.$$ So $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c.$$ Now you can eliminate $a+b+c$ from your first inequality.

Another way to do this would be the following (I'm doing Liu Gang's suggested generalization):

We have to show $$\frac{a^{n+1}}{b^n} + \frac{b^{n+1}}{c^n} + \frac{c^{n+1}}{a^n} - \frac{a^n}{b^{n-1}} - \frac{b^n}{c^{n-1}} - \frac{c^n}{a^{n-1}} \ge 0.$$ The left hand side equals $$\frac{a^n(a - b)}{b^n} + \frac{b^n(b-c)}{c^n} + \frac{c^n(c-a)}{a^n},$$ and therefore it is enough to show that $$c^n a^{2n} (a-b) + a^n b^{2n} (b-c) + b^n c^{2n} (c-a) \ge 0.$$ Because the inequality is cyclic, we can assume that either $a \ge b \ge c$ or $a \ge c \ge b$.

In the first case we have $c^n a^{2n} \ge b^n c^{2n}$ and $a^n b^{2n} \ge b^n c^{2n}$, so we get that the LHS is $\ge b^n c^{2n} (a - b + b - c + c - a) = 0$.

In the second case we have $a^n b^{2n} \le c^n a^{2n}$ and $b^n c^{2n} \le c^n a^{2n}$, so we get that the LHS is $\ge c^n a^{2n} (a - b + b - c + c- a) = 0$.

This proves the claim.