Prove that $\sum\limits_{cyc}\sqrt{a+11bc+6}\geq9\sqrt2.$

Question

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.

Answer 1

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.