Prove $\sum \limits_{cyc} \frac{1+b^2+c^4}{a^+b^2+c^3}\geq 3$ 
If $a,b,c$ are positive real numbers, prove $$\sum \limits_{cyc} \frac{1+b^2+c^4}{a+b^2+c^3}\geq 3$$
  Additional info: We should only use Cauchy (preferred to used at least once and more than AM-GM) and AM-GM. We are not allowed to use induction.

Things I have done so far: My idea is about separating question into $3$ inequality and proving them one by one and then sum all of them to prove question inequality.So$$\sum \limits_{cyc} \frac{1}{a+b^2+c^3} \geq 1$$$$\sum \limits_{cyc} \frac{b^2}{a+b^2+c^3} \geq 1$$
$$\sum \limits_{cyc} \frac{c^4}{a+b^2+c^3} \geq 1$$
UPDATE
As Macavity pointed in comments, the separating idea is not useful because first inequality does not hold.So any hint for starting step is appreciated. 

Using Cauchy-Schwarz, note
$$LHS = \sum_{cyc} \frac{(1+b^2+c^4)(a+b^2+c^3)}{(a+b^2+c^3)^2} \ge  \sum_{cyc} \frac{(1+b^2+c^4)(a+b^2+c^3)}{(a^2+b^2+c^2)(1+b^2+c^4)} = \sum_{cyc} \frac{a+b^2+c^3}{a^2+b^2+c^2}$$
As the denominators of the cyclic sum are now the same, it only remains to show that
$$\sum_{cyc} (a + b^2+c^3) \ge 3(a^2+b^2+c^2) \iff \sum_{cyc} a + \sum_{cyc} a^3 \ge 2\sum_{cyc} a^2$$
which follows from AM-GM as $a+a^3 \ge 2a^2$.
