TITLE: 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.
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: Another way:
Let $a\geq b\geq c$.
Thus,
$$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}-\frac{3}{2}=\sum_{cyc}\frac{4a^2}{2\sqrt{(b+c)^2(4b^2-4bc+4c^2)}}-\frac{3}{2}\geq$$
$$\geq\sum_{cyc}\left(\frac{4a^2}{(b+c)^2+4b^2-4bc+4c^2}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{8a^2-5b^2-5c^2+2bc}{5b^2-2bc+5c^2}=$$
$$=\frac{1}{2}\sum_{cyc}\frac{(a-b)(4a+5b-c)-(c-a)(4a+5c-b)}{5b^2-2bc+5c^2}=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{4a+5b-c}{5b^2-2bc+5c^2}-\frac{4b+5a-c}{5a^2-2ac+5c^2}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(20a^2+45ab+20b^2-13ac-13bc-3c^2)}{(5a^2-2ac+5c^2)(5b^2-2bc+5c^2)}\geq$$
$$\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(16(a+b)^2-13ac-13bc-3c^2)}{(5a^2-2ac+5c^2)(5b^2-2bc+5c^2)}=$$
$$=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2(a+b-c)(16a+16b+3c)}{(5a^2-2ac+5c^2)(5b^2-2bc+5c^2)}\geq$$
$$\geq\frac{1}{2}\left(\frac{(a-c)^2(a+c-b)(16a+16c+3b)}{(5a^2-2ab+5b^2)(5b^2-2bc+5c^2)}+\frac{(b-c)^2(b+c-a)(16b+16c+3a)}{(5a^2-2ab+5b^2)(5a^2-2ac+5c^2)}\right)\geq$$
$$\geq\frac{1}{2}\left(\frac{(b-c)^2(a-b)(16b+16c+3a)}{(5a^2-2ab+5b^2)(5a^2-2ac+5c^2)}-\frac{(b-c)^2(a-b)(16b+16c+3a)}{(5a^2-2ab+5b^2)(5a^2-2ac+5c^2)}\right)=0.$$