By Holder $$\sum\limits_{cyc}\sqrt{\tfrac{a^3}{a^2+8b^2}}=\sqrt{\tfrac{\left(\sum\limits_{cyc}\sqrt{\tfrac{a^3}{a^2+8b^2}}\right)^2\sum\limits_{cyc}(a^2+8b^2)(c^2+ca+2a^2)^3}{\sum\limits_{cyc}(a^2+8b^2)(c^2+ca+2a^2)^3}}\geq\sqrt{\tfrac{\left(\sum\limits_{cyc}(2a^3+a^2b+a^2c)\right)^3}{\sum\limits_{cyc}(a^2+8b^2)(c^2+ca+2a^2)^3}}$$ and it's enough to prove that
$$(a^2+b^2+c^2)\left(\sum\limits_{cyc}(2a^3+a^2b+a^2c)\right)^3\geq(a^3+b^3+c^3)\sum\limits_{cyc}(a^2+8b^2)(c^2+ca+2a^2)^3$$ or
$$\sum_{cyc}(12a^{10}b-39a^9b^2+38a^8b^3+28a^8c^3+42a^7b^4+30a^7c^4-8a^6b^5+41a^6c^5)+$$
$$+abc\sum_{cyc}(12a^8-57a^7b+15a^7c+42a^6b^2+54a^6c^2+57a^5b^3+57a^5c^3-36a^4b^4)+$$
$$+a^2b^2c^2\sum_{cyc}(-138a^5-9a^4b-48a^4c-108a^3b^2-180a^3c^2+60a^3bc+135a^2b^2c)\geq0,$$ which is true by BW.
My second proof using Holder inequality.
By Holder inequality, we have
\begin{align*}
&\left(\sum_{\mathrm{cyc}}\frac{1}{\sqrt{2a + bc}}\right)^2 \sum_{\mathrm{cyc}} (2a + bc)(2ab + 2bc + 2ca + b\sqrt 2 + c\sqrt 2)^3\\
\ge{}& \left(6ab + 6bc + 6ca + 2a\sqrt 2 + 2b\sqrt{2} + 2c\sqrt{2}\right)^3.
\end{align*}
It suffices to prove that
\begin{align*}
&\left(6ab + 6bc + 6ca + 2a\sqrt 2 + 2b\sqrt{2} + 2c\sqrt{2}\right)^3\\
\ge{}& (1 + \sqrt 2)^2\sum_{\mathrm{cyc}} (2a + bc)(2ab + 2bc + 2ca + b\sqrt 2 + c\sqrt 2)^3. \tag{1}
\end{align*}
We use the pqr method.
Let $p = a + b + c, q = ab + bc + ca, r = abc$. Using $p^2 \ge 3q = 3$, we have $p \ge \sqrt 3$.
(1) is written as
$$c_1 r + c_0 \ge 0\tag{2}$$
where
\begin{align*}
c_1 &= 42\,\sqrt {2}{p}^{2}+60\,\sqrt {2}p+56\,{p}^{2}-48\,\sqrt {2}+
100\,p-88, \\
c_0 &= 10\,\sqrt {2}{p}^{3}-36\,\sqrt {2}{p}^{2}-8\,{p}^{
3}+100\,\sqrt {2}p+92\,{p}^{2}-136\,\sqrt {2}-168\,p+32.
\end{align*}
We have $c_1 \ge 0.$
If $p^2 \ge 4q = 4$, we have $c_0 \ge 0$. Thus, (2) is true.
If $p^2 < 4q = 4$, using degree three Schur $r \ge \frac{4pq - p^3}{9} = \frac{4p - p^3}{9}$, we have
$$c_1r + c_0 \ge c_1 \cdot \frac{4p - p^3}{9} + c_0 \ge 0.$$
We are done.
Best Answer
Your way leads to a wrong inequality.
Try $(a,b,c)=(3,1,0).$
My solution:
Since by C-S $$\sqrt{b^2+7bc+c^2}=\frac{\sqrt{\left((b-c)^2+4bc\right)\left((b-c)^2+9bc\right)}}{b+c}\geq$$ $$\geq\frac{(b-c)^2+6bc}{b+c}=\frac{b^2+4bc+c^2}{b+c},$$ it's enough to prove that: $$\sum_{cyc}\frac{a(b+c)}{b^2+4bc+c^2}\leq\frac{a^2+b^2+c^2}{ab+ac+bc}$$ or $$abc\sum_{sym}(a^5+7a^4b-4a^3b^2+8a^3bc-12a^2b^2c)\geq0,$$ which is true by Muirhead.