Prove the inequality $\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32$ where $a,b,c$ are positive reals. 
Let $a,b,c$ be positive real numbers, prove that $$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32.$$

The problem is from an inequality handout. Here is my attempt to solve the problem:
I first rewrote the inequality as $$\sum_{cyc}\frac{a^2}{(b+c)\sqrt{b^2-bc+c^2}}\geq\frac 32.$$
Here I noticed that the inequality has some similarities with Nesbitt's inequality, because that says $$\sum_{cyc}\frac{a}{b+c}\geq\frac 32.$$
But I don't know how to use this to reach the given inequality.
I also used Cauchy-Schwarz and got $$\sum_{cyc}\frac{a^2}{(b+c)\sqrt{b^2-bc+c^2}}\geq\frac{(a+b+c)^2}{\sum_{cyc}(b+c)\sqrt{b^2-bc+c^2}}.$$
But this also makes the problem too complicated that involves too many square roots. I think the square root on the denominator should be removed first to prove the inequality. But I am unable to do that.
So, I need a solution to the problem. Also it would be helpful for me if the answerer provides some motivation for the solution.

Another way.
By Holder $$\left(\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\right)^2\sum_{cyc}a^2(b+c)(b^3+c^3)\geq(a^2+b^2+c^2)^3$$ and it's enough to prove that:
$$4(a^2+b^2+c^2)^3\geq9\sum_{cyc}(a^2b+a^2c)(b^3+c^3)$$ or
$$\sum_{cyc}(4a^6+3a^4b^2+3a^4c^2-9a^3b^2c-9a^3c^2b+8a^2b^2c^2)\geq0,$$ which is true by Schur, AM-GM, Schur and Muirhead:
$$\sum_{cyc}(4a^6+3a^4b^2+3a^4c^2-9a^3b^2c-9a^3c^2b+8a^2b^2c^2)\geq$$
$$\geq\sum_{cyc}(7a^4b^2+7a^4c^2-9a^3b^2c-9a^3c^2b+4a^2b^2c^2)\geq$$
$$\geq\sum_{cyc}(14a^4bc-9a^3b^2c-9a^3c^2b+4a^2b^2c^2)=$$
$$=9abc\sum_{cyc}(a^3-a^2b-a^2c+abc)+5abc\sum_{cyc}(a^3-abc)\geq0.$$
