If $ab+ac+bc=3$ so $\sum\limits_{cyc}\sqrt{a^2+b^2+7bc}\geq9$. Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$. Prove that:
$$\sqrt{a^2+b^2+7bc}+\sqrt{b^2+c^2+7ca}+\sqrt{c^2+a^2+7ab}\geq9$$
The equality occurs also for $a=2$, $b=\frac{3}{2}$ and $c=0$.
I tried the following Holder.
$$\left(\sum\limits_{cyc}\sqrt{a^2+b^2+7bc}\right)^2\sum_{cyc}(a^2+b^2+7bc)^2(ka+mb+c)^3\geq$$
$$\geq\left(\sum_{cyc}(a^2+b^2+7bc)(ka+mb+c)\right)^3.$$ 
Thus, it remains to prove that 
$$\left(\sum_{cyc}(a^2+b^2+7bc)(ka+mb+c)\right)^3\geq9(ab+ac+bc)^2\sum_{cyc}(a^2+b^2+7bc)^2(ka+mb+c)^3,$$
but I did not find a non-negative values of $k$ and $m$, for which the last inequality would be true.

Proof
By Holder, we have
\begin{align}
&\Big(\sum_{\mathrm{cyc}} \sqrt{a^2+b^2+7bc}\Big)^2\sum_{\mathrm{cyc}} (a^2+b^2+7bc)^2
(6ab + 3bc + 10ca)^3\\
\ge \ &
\Big(\sum_{\mathrm{cyc}} (a^2+b^2+7bc)(6ab+3bc+10ca)\Big)^3.
\end{align}
Thus, it suffices to prove that
\begin{align}
&\Big(\sum_{\mathrm{cyc}} (a^2+b^2+7bc)(6ab+3bc+10ca)\Big)^3\\
\ge \ & 27(ab+bc+ca)\sum_{\mathrm{cyc}} (a^2+b^2+7bc)^2
(6ab + 3bc + 10ca)^3.
\end{align}
Let $f(a,b,c) = \mathrm{LHS}-\mathrm{RHS}$.
The Buffalo Way works. WLOG, assume that $c = \min(a,b,c)$. There are two possible cases:
1) $c\le b \le a$: Let $b = c + s,\ a = c+s+t; \ s,t\ge 0$. Then
$$f(c+s+t, c+s, c) = g(c,s,t) + s^3(352s^4+1068s^3t+1269s^2t^2+378st^3+81t^4)(s-3t)^2(s+t)^3$$
where $g(c,s,t)$ is a polynomial with non-negative coefficients. True.
2) $c\le a\le b$: Let $a=c+s, \ b = c+s+t; \ s,t\ge 0$. Then
$f(c+s, c+s+t, c)$ is a polynomial in $c, s, t$ with non-negative coefficients. True.
We are done.
