prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$ 
If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$
  Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.

Things I have done so far:
Using Cauchy inequalities I can write:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})(b+c+a) \geq (a+b+c)^2$$
$a+b+c=1$,So:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})\geq 1$$

In this case you want to have quadratic terms in the RHS, so use Cauchy in the following form:
$$\left(\sum_{cyc} \frac{a^2}b \right) \left(\sum_{cyc} a^2b \right) \ge \left(a^2+b^2+c^2\right)^2 \tag{1}$$
Then it is enough to show that
$$a^2+b^2+c^2 \ge 3\sum_{cyc} a^2b   \tag{2}$$
Homogenising, 
$$\iff \left( a^2+b^2+c^2\right) (a+b+c) \ge 3\sum_{cyc} a^2b  $$
$$\iff \sum_{cyc} a^3+ \sum_{cyc} ab^2 \ge 2\sum_{cyc} a^2b  $$
which follows from AM-GM as $a^3+ab^2 \ge 2a^2b$.
