Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$ 
If $a$, $b$ and $c$ are positive real numbers, prove that:
   $$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$

Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have tried so far:
Using Cauchy inequality I can write:$$\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)(a+b+c) \geq \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2$$
but I can't continue this.I tried expanded form:$$\sum \limits_{cyc} \frac{a^5c^2}{a^2b^2c^2} \geq \sum \limits_{cyc}  \frac{a^3c}{abc}$$
Which proceeds me to this Cauchy:$$\sum \limits_{cyc} \frac{a^5c^2}{abc}\sum \limits_{cyc}a(abc)\geq \left(\sum \limits_{cyc}a^3c\right)^2$$
I can't continue this one too.
The main Challenge is $3$ fraction on both sides which all of them have different denominator.and it seems like using Cauchy from first step won't leads to anything good.  
 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.
A: 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.
A: $$\sum_{cyc}\frac{a^3}{b^2}-\sum_{cyc}\frac{a^2}{b}=\sum_{cyc}\left(\frac{a^3}{b^2}-\frac{a^2}{b}\right)=$$
$$=\sum_{cyc}\left(\frac{a^2(a-b)}{b^2}-(a-b)\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{b^2}\geq0.$$
Done!
A: By AM–GM, we have
$$14\frac{a^3}{b^2}+3\frac{b^3}{c^2}+2\frac{c^3}{a^2}\geq 19\frac{a^2}{b}\quad (1)$$
$$2\frac{a^3}{b^2}+14\frac{b^3}{c^2}+3\frac{c^3}{a^2}\geq 19\frac{b^2}{c}\quad (2)$$
$$3\frac{a^3}{b^2}+2\frac{b^3}{c^2}+14\frac{c^3}{a^2}\geq 19\frac{c^2}{a}\quad (3)$$
Add all three equations together and conclude.
A: By AM-GM, we have:
$$\frac{a^3}{b^2} + a\ge \frac{2a^2}{b}\Rightarrow \frac{a^3}{b^2}\geq \frac{2a^2}{b} -a$$
Similar: $$\frac{b^3}{c^2}\geq \frac{2b^2}{c} -b$$
$$\frac{c^3}{a^2}\geq \frac{2c^2}{a} -c$$
Now adding them together and using C-S, we have:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2} + \frac{c^3}{a^2}$$
$$\geq (\frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a})+(\frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a}) - (a+b+c)$$
$$\geq (\frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a})+\frac{(a+b+c)^2}{a+b+c} - (a+b+c)$$
$$= \frac{a^2}{b} + \frac{b^2}{c}+\frac{c^2}{a}$$
Equality holds when $a=b=c$
P.s: I don't think it's a nice solution:(
