Prove that $\sum\limits_{cyc}\sqrt{a+11bc+6}\geq9\sqrt2.$ 
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc+abc=4.$ Prove that:
  $$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}+\sqrt{c+11ab+6}\geq 9\sqrt2.$$

The equality occurs for $(a,b,c)=(1,1,1)$ and again for $(a,b,c)=(2,2,0)$ and for the cyclic permutations of the last.
I tried Holder:
$$\sum_{cyc}\sqrt{a+11bc+6}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{a+11bc+6}\right)^2\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}{\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}}\geq$$
$$\geq \sqrt{\frac{\left(\sum\limits_{cyc}(a+11bc+6)(3a+4)\right)^3}{\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}}$$ and it's enough to prove that
$$\left(\sum\limits_{cyc}(a+11bc+6)(3a+4)\right)^3\geq162\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3,$$
which is true for $(a,b,c)=(2,2,0)$, $(a,b,c)=(1,1,1)$, but it's wrong for $(a,b,c)=\left(8,\frac{1}{2},0\right).$ 

Thanks to  River Li for this counterexample.

Also, I tried a substitution $a=\frac{2x}{y+z},$ $b=\frac{2y}{x+z}$, $c=\frac{2z}{x+y}$ and SOS, but it seems very complicated.
Also, I tried the following estimation. By Minkowcki:
$$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}\geq\sqrt{(\sqrt{a}+\sqrt{b})^2(1+11c)+24}.$$
Now, for $c=\min\{a,b,c\}$ it's enough to prove that
$$\sqrt{(\sqrt{a}+\sqrt{b})^2(1+11c)+24}+\sqrt{c+11ab+6}\geq9\sqrt2,$$ which not so helps.
Also, LM does not help.
Thank you! 
Update
Also, there is the following.
We need to prove that:
$$\sum_{cyc}\sqrt{\frac{2x}{y+z}+\frac{44yz}{(x+y)(x+z)}+6}\geq9\sqrt2$$ or
$$\sum_{cyc}\sqrt{\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3}\geq9,$$ where $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0.$
Now, by Holder
$$\left(\sum_{cyc}\sqrt{\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3}\right)^2\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(kx^2+y^2+z^2+myz+nxy+nxz)^3\geq$$
$$\geq\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(kx^2+y^2+z^2+myz+nxy+nxz)\right)^3,$$ where $k$, $m$ and $n$ are reals such that the expression $kx^2+y^2+z^2+myz+nxy+nxz$ 
is non-negative for all non-negatives $x$, $y$ and $z$.
Thus, it's enough to choose values of $k$, $m$ and $n$ for which the following inequality is true.
$$\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(kx^2+y^2+z^2+myz+nxy+nxz)\right)^3\geq$$
$$\geq81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(kx^2+y^2+z^2+myz+nxy+nxz)^3$$
From the equality case we can get that should be $$2k-5m+2n=8.$$
For $k=1$, $m=0$ and $n=3$ we need to prove that:
$$\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(x^2+y^2+z^2+3xy+3xz)\right)^3\geq$$
$$\geq81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(x^2+y^2+z^2+3xy+3xz)^3,$$
which is true for $y=z$ and it's true for $z=0$, but I have no a proof for all non-negative variables.

Michael Rozenberg actually gave a proof. I do a little bit by the Buffalo Way to prove that
\begin{align}
&\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(x^2+y^2+z^2+3xy+3xz)\right)^3\\
\geq\ & 81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(x^2+y^2+z^2+3xy+3xz)^3.
\end{align}
It suffices to prove that $f(x,y,z)\ge 0$ where $f(x,y,z)$ is a polynomial (a long expression).
WLOG, assume that $z = \min(x,y,z).$ There are two possible cases:
1) If $z \le y \le x$, let $y=z+s, \ x = z+s+t; \ s,t \ge 0$.
Note that $f(z+s+t, z+s, z)$ is a polynomial in $z, s, t$ with non-negative coefficients.
It is true. 
2) If $z \le x\le y$, let $x = z+s, \ y = z+s+t; \ s,t \ge 0.$
Note that $f(z+s, z+s+t, z)$ is a polynomial in $z, s, t$ with non-negative coefficients.
It is true. We are done.
