$(xy+yz+zx)(1+6xyz) \geq 11xyz$ I've came across this inequality:

Let $x>$, $y>0$ and $z > 0$ with  $x+y+z=1$.
   Prove that
  $$(xy+yz+zx)(1+6xyz) \geq 11xyz.$$

I don't know where to take it from, I've tried means inequality, but it didn't help me. 
Some hints would be great!
 A: It's $$6xyz(xy+xz+yz)+(xy+xz+yz)(x+y+z)^3\geq11xyz(x+y+z)^2$$ or
$$\sum_{sym}(x^4y+3x^3y^2-2x^3yz-2x^2y^2z)\geq0,$$
which is true by Muirhead.
Also, we can prove the last inequality by AM-GM.
Indeed, $$\sum_{cyc}x^4y=\frac{1}{13}\sum_{cyc}(9x^4y+y^4z+3z^4x)\geq$$
$$\geq\frac{1}{13}\cdot13\sum_{cyc}\sqrt[13]{\left(x^4y\right)^9\left(y^4z\right)\left(z^4x\right)^3}=\sum_{cyc}x^3yz;$$
$$\sum_{cyc}x^4z=\frac{1}{13}\sum_{cyc}(9x^4z+y^4x+3z^4y)\geq$$
$$\geq\frac{1}{13}\cdot13\sum_{cyc}\sqrt[13]{\left(x^4y\right)^9\left(y^4z\right)\left(z^4x\right)^3}=\sum_{cyc}x^3yz;$$
$$\sum_{cyc}(x^3y^2+x^3z^2)\geq2\sum_{cyc}\sqrt{x^3y^2x^3z^2}=2\sum_{cyc}x^3yz$$
and
$$\sum_{cyc}x^3yz=\frac{1}{2}xyz\sum_{cyc}(x^2+y^2)\geq\frac{1}{2}xyz\cdot2\sum_{cyc}\sqrt{x^2y^2}=xyz\sum_{cyc}xy=\sum_{cyc}x^2y^2z.$$
Also, there are easy way by SOS:
$$\sum_{sym}(x^4y+3x^3y^2-2x^3yz-2x^2y^2z)=$$
$$=\sum_{cyc}(x^4y+x^4z+3x^3y^2+3x^3z^2-4x^3yz-4x^2y^2z)=$$
$$=\sum_{cyc}(x^4y+xy^4-x^3y^2-x^2y^3)+$$
$$+2\sum_{cyc}(z^3x^2+z^3y^2-2z^3xy+x^3z^2+y^3z^2-x^2z^2y-y^2z^2x)=$$
$$=\sum_{cyc}(x-y)^2(xy(x+y)+2z^3+2z^2(x+y))\geq0.$$
There are also  easy proofs by uvw and by Rearrangement, but Muirhead it's the best, I think.
A: $$p= a+b+c=1, \ q=ab+bc+ca,\ r= abc\le \dfrac {1}{27} $$
$$ q^2\ge 3pr \ , q (1+6r)\ge \sqrt {3r}(1+6r)\ge 11r $$
$$(1-27r)(3-4r)\ge 0 $$
