Prove the inequality $\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}\ge 1$ 
Let $a,b,c>0$ and such $a+b+c=3$,show that
$$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\ge 1\tag{1}$$

I tried using Holder's inequality to solve it:
$$\sum_{cyc}\dfrac{a^3}{b\sqrt{a^3+8}}\sum b\sum \sqrt{a^3+8}\ge (a+b+c)^3$$
But the following is not right
$$\sum\sqrt{a^3+8}\le 9$$
so please help me prove $(1)$
 A: By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^3}{b\sqrt{a^3+8}}=\sum_{cyc}\frac{2a^3}{b\cdot2\sqrt{(a+2)(a^2-2a+4)}}\geq\sum_{cyc}\frac{2a^3}{b(a+2+a^2-2a+4)}=$$
$$=\sum_{cyc}\frac{2a^4}{ab(a^2-a+6)}\geq\frac{2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}ab(a^2-a+6)}=$$
$$=\frac{6(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(3a^3b-a^2b(a+b+c)+2ab(a+b+c)^2)}=\frac{6(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^3b+2a^3c+3a^2b^2+9a^2bc)}$$ and it's enough to prove that:
$$\frac{6(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^3b+2a^3c+3a^2b^2+9a^2bc)}\geq1,$$
which is true by SOS and Tangent Line methods:
$$6(a^2+b^2+c^2)^2-\sum\limits_{cyc}(4a^3b+2a^3c+3a^2b^2+9a^2bc)=$$
$$\sum\limits_{cyc}(6a^4-4a^3b-2a^3c+9a^2b^2-9a^2bc)=$$
$$=2\sum_{cyc}(3a^4-2a^3b-ab^3)+\frac{9}{2}\sum_{cyc}(c^2a^2-2c^2ab+c^2b^2)=$$
$$=2\sum_{cyc}a(a-b)(3a^2+ab+b^2)+\frac{9}{2}\sum_{cyc}c^2(a-b)^2=$$
$$=2\sum_{cyc}\left((a-b)(3a^3+a^2b+ab^2)-\frac{5}{4}(a^4-b^4)\right)+\frac{9}{2}\sum_{cyc}c^2(a-b)^2=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2(7a^2+6ab+5b^2)+\frac{9}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
