<s>[INST] 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.  
 [/INST] 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:(
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