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.  

$$\sum_{cyc}\frac{a^3}{b^2}-\sum_{cyc}\frac{a^2}{b}=\sum_{cyc}\left(\frac{a^3}{b^2}-\frac{a^2}{b}\right)=$$
$$=\sum_{cyc}\left(\frac{a^2(a-b)}{b^2}-(a-b)\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{b^2}\geq0.$$
Done!
