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 $(1)$