If $a$, $b$, $c$ are sides of a triangle, prove $2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$ 
$a$, $b$, $c$ are sides of a triangle, prove:
  $$2(a+b+c)(a^2+b^2+c^2)\geq3(a^3+b^3+c^3+3abc)$$

What I have tried:
$$
⇔2\sum (a+b)ab\geq \sum a^3+9abc
$$
so I can't use 
$$\sum a^3+3abc\geq \sum ab(a+b)$$

Also, we can use SOS here.
Indeed, by your work we need to prove that 
$$\sum_{cyc}(2a^2b+2a^2c-a^3-3abc)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, since $$3b-a-c=b+c-a+2b-2c>0,$$ we obtain:
$$\sum_{cyc}(2a^2b+2a^2c-a^3-3abc)=\sum_{cyc}(2a^2b+2a^2c-4abc-a^3+abc)=$$
$$=\sum_{cyc}\left(2c(a-b)^2-\frac{1}{2}(a+b+c)(a-b)^2\right)=\frac{1}{2}\sum_{cyc}(a-b)^2(3c-a-b)\geq$$
$$\geq\frac{1}{2}((a-b)^2(3c-a-b)+(a-c)^2(3b-a-c))\geq$$
$$\geq\frac{1}{2}((a-b)^2(3c-a-b)+(a-b)^2(3b-a-c))=(a-b)^2(b+c-a)\geq0.$$
