For $a=b=c=\frac{1}{3}$ we obtain $m\geq27$. But $27(a^3+b^3+c^3)\geq6(a^2+b^2+c^2)(a+b+c)+(a+b+c)^3$ is obviously true.

$0<a,b,c<1.$ $(3a-1)^2(9a+4)\ge 0\Leftrightarrow 27a^3\ge 6a^2+5a-\frac{4}{3}$ $\Rightarrow 27(a^3+b^3+c^3) \ge 6(a^2+b^2+c^2)+5(a+b+c)-4= 6(a^2+b^2+c^2)+1.$

Let $A= a^3+b^3+c^3 $ We have:: (1) $A \ge \frac{1}{9}$ (2) $a^2+b^2+c^2 \le 3*(A/3)^{\frac{2}{3}} $ (3) $(A/3)^{\frac{2}{3}} \le A$ So: $ 6(a^2+b^2+c^2)+1 \le 18A+9A=27A$