Prove that: $\sum_{\text {sym}} a^3+3\sum_{\text{sym}} a^2b\ge 18$ Let $a,b,c$ be the sides of triangle, such that $abc=1$, then prove that:

$$\sum_{\text{sym}} a^3+3\sum_{\text{sym}} a^2b\ge 18.$$
Find at least one case where equality is possible.

My attempts:
I tried the well-known formula with triangle inequalities
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)>0$$
But, this formula didn't help.

$$3(a^2b+c^2b+b^2c+a^2c+c^2a+b^2a)=$$ $$=3abc((a/c+c/a)+(b/a+a/b)+(c/b+b/c))\ge$$ $$\ge 3(6)=18$$ because $abc=1$ and because if $x\in\{a/c,b/a,c/b\}$ then $x+1/x=(\sqrt x\,-1/\sqrt x)^2+2\ge 2.$
