how to prove this inequality $(ab+bc+ac)^2 ≥ 3abc(a+b+c)$ 
Prove that if $a,b,c$ are non-negative real numbers, then
  $(ab + bc + ca)^2 \geq 3abc(a+b+c)$.

I tried to compute from $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$.

$a,b,c$ can be any real numbers.
Let $x=ab$, $y=bc$, $z=ca$. Then your inequality is $$(x+y+z)^2\ge 3(xy+yz+zx),$$
which is true, because $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
and the inequality is equivalent to $$x^2+y^2+z^2\ge xy+yz+zx,$$
which is true, because, as you said, it's equivalent to $$\frac{1}{2}\left((x-y)^2+(y-z)^2+(z-x)^2\right)\ge 0,$$
which is true; or you can use Rearrangement Inequality.
Equality holds if and only if $x=y=z$, i.e. iff either at least two of $a,b,c$ are equal to $0$ or $abc\neq 0$ and $a=b=c$.
