Very strong inequality Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a^3+b^3+c^3+3abc\geq(a+b-c)\sqrt{ab(a+c)(b+c)}+(a+c-b)\sqrt{ac(a+b)(b+c)}+(b+c-a)\sqrt{bc(a+b)(a+c)}$$
I have a proof, but my proof is long. Maybe there is something simple?
Thanks! 

Here is a brutal force method and I am not sure it is shorter than op:
it is easy to verify if $abc=0$, the inequality holds.
then suppose $a,b,c$ are sides of a triangle, then we can take the advantage of Thunderstruck's proof.
if $a,b,c$ are not sides of a triangle,WLOG, $a \ge b+c$, now we prove :
LHS $>(a+b-c)\sqrt{ab(a+c)(b+c)}+(a+c-b)\sqrt{ac(a+b)(b+c)}$
we have:
$((a+b-c)^2+(a+c-b)^2)(ab(a+c)(b+c)+ac(a+b)(b+c))=2a(a^2+(b-c)^2)(b+c)(ab+ac+2bc) \ge  ((a+b-c)\sqrt{ab(a+c)(b+c)}+(a+c-b)\sqrt{ac(a+b)(b+c)})^2$
$a=b+c+x \implies x\ge0 $
$(a^3+b^3+c^3+3abc)^2-2a(a^2+(b-c)^2)(b+c)(ab+ac+2bc)=x^6+6cx^5+6bx^5+13c^2x^4+32bcx^4+13b^2x^4+14c^3x^3+56bc^2x^3+56b^2cx^3+14b^3x^3+7c^4x^2+42bc^3x^2+79b^2c^2x^2+42b^3cx^2+7b^4x^2+12bc^4x+52b^2c^3x+52b^3c^2x+12b^4cx+16b^2c^4+32b^3c^3+16b^4c^2>0$
