The task is homogenius and symmetric by the $b,c.$
Let WLOG
$$a=1,\quad s = \dfrac{b+c}2,\quad t=\left(\dfrac{b-c}{b+c}\right)^2,\tag1$$
then
\begin{align}
&b+c = 2s,\quad bc = s^2(1-t),\quad b^2+c^2=2s^2(1+t),\\[4pt]
&b^4+c^4 = (b^2+c^2)^2 - 2(bc)^2 = 2s^4(t^2+6t+1),\\[4pt]
&\dfrac{b}{1+c}+\dfrac{c}{1+b} = \dfrac{b+c+b^2+c^2}{1+b+c+bc} = \dfrac{2s+2s^2(1+t)}{1+2s+s^2(1-t)},
\end{align}
and the issue task transforms to
$$\dfrac1{s\sqrt[4]{t^2+6t+1}} + \dfrac{4s+4s^2(1+t)}{1+s+s^2(1-t)} \ge 3, \quad s > 0,\quad 1\ge t\ge0,$$
or
$$\dfrac1{s\sqrt[4]{t^2+6t+1}} + 4\dfrac{(1+s)(1+2s)}{(1+s)^2-s^2t} \ge 7, \quad s > 0,\quad 1\ge t\ge0,$$
or
$$\dfrac{v-1}{\sqrt[4]{t^2+6t+1}} + 4\dfrac{v(v+1)}{v^2-t} \ge 7, \quad v > 1,\quad 1\ge t\ge0, \tag2,$$
where
$$v=\dfrac1s+1.\tag3$$
Then, using the evident inequality
$$\sqrt{1\pm x}\le 1\pm \dfrac x2,$$
easy to get
$$\sqrt{1+6t^2+t^4} = (1+3t)\sqrt{1-\dfrac {8t^2}{(1+3t)^2}}\le(1+3t)\left(1-\dfrac{4t^2}{(1+3t)^2}\right) = \dfrac{1+6t+5t^2}{1+3t},$$
$$\sqrt{1+6t+5t^2} = (1+3t)\sqrt{1-\dfrac {4t^2}{(1+3t)^2}}\le\dfrac{1+3t}{1+\dfrac{2t^2}{(1+3t)^2}} = \dfrac{(1+3t)^3}{1+6t+11t^2}.$$
This allows to prove the stronger inequality
$${1+6t+11t^2\over (\sqrt{1+3t\,})^5}(v-1) + 4\dfrac{v(v+1)}{v^2-t} \ge 7, \quad v > 1,\quad 1\ge t\ge0.$$
Substitution
$$z=\sqrt{1+3t}\tag4$$
leads to the equivalent task
$${2-4z^2+11z^4\over 9z^5}(v-1)+12{v(v+1)\over 3v^2 + 1-z^2} \ge 7,\quad z\ge 1,\quad 2 \ge z\ge 1,$$
or
$$f(z,v) \ge 0,\quad v>1,\quad 2 \ge z\ge1,\tag5$$
wherein
\begin{align}
f(z,v) = (3v^2+1-z^2)((v-1)(11z^4-4z^2+2)-63z^5)+108z^5(v^2+v).\tag6
\end{align}
The highest value of $f(x)$ can be achieved on its stationary points or the area bounds.
The stationary points
The stationary points of $f(z,v)$ can be obtained from the conditions $f'_v=\dfrac1{3z}f'_v=0,$ and that leads to the system
\begin{cases}
(99z^4-36z^2+18)v^2+(-162z^5-66z^4+24z^2-12)v-11z^6+108z^5+15z^4-6z^2+2 = 0\\[4pt]
(44z^2-8)v^3+(-135z^3-44z^2+8)v^2+(-22z^4+180z^3+20z^2-4)v+147z^5+22z^4-105z^3-20z^2+4=0\\[4pt]
v>1,\quad 2 \ge z\ge1,\tag7
\end{cases}
(see also Wolfram Alpha), with the solutions
$$\begin{pmatrix}v\\z\\f(v,z)\end{pmatrix}
=\left\{
\begin{pmatrix}2\\1\\0\end{pmatrix},
\begin{pmatrix}\approx 1.713140\\\approx 0.883476\\ \approx 0.0969191\end{pmatrix}
\right\}\tag8$$
which can be obtained using the polynomial reduction way.
Taking in account the calculations accuracy, this means that $f(v,z)\ge 0$ in the stationary points.
The bounds
If $\mathbf{v=1,}$ then $f(1,z) = 9z^5(7z^2-4)\ge 0\text{ if } z\in[1,2].$
If $\mathbf{z=1},$ then $f(v, 1) = 27v(v-2)^2 \ge 0 \text{ if } v\in(1,\infty).$
If $\mathbf{z=2},$ then
$f(v,2) = 54 (11 - 3 v)^2 (v + 1) \ge 0,\text{ if } v\in(1,\infty).$
These mean that $f(v,z)\ge 0$ on the area $v\in(1, \infty), \quad z\in [1,2].$
Proved.
Update 08.05.2018
The polynomial reduction way
Taking in account the elder coefficients of the derivatives, let us consider the system
$$\dfrac{df}{dv} = 4v(11z^2-2)\dfrac{df}{dv} - \dfrac{-33z^4+12z^2-6}{z}\dfrac{df}{dz} = 0,$$
or
$$-6(27z^5+11z^4-4z^2+2)v+9(11z^4-4z^2+2)v^2 = 11z^6-108z^5-15z^4+6z^2-2\\
2(847z^8-6534z^7-1012z^6+2808z^5+564z^4-1620z^3-184z^2+28)v+ 3(2079z^7+484z^6-1188z^5-264z^4+810z^3+120z^2-16)v^2 = 9(1617z^9+242z^8-1743z^7-308z^6+714z^5+168z^4-210z^3-56z^2+8)
$$
$$\begin{cases}
-6(27z^5+11z^4-4z^2+2)v + 9(11z^4-4z^2+2)v^2=11z^6-108z^5-15z^4+6z^2-2\\[4pt]
2(847z^8-6534z^7-1012z^6+2808z^5+564z^4-1620z^3-184z^2+28)v\\
+3(2079z^7+484z^6-1188z^5-264z^4+810z^3+120z^2-16)v^2\\
=9(1617z^9+242z^8-1743z^7-308z^6+714z^5+168z^4-210z^3-56z^2+8)\\[4pt]
v>1,\quad 2 \ge z\ge1,
\end{cases}\tag9$$
(see also Wolfram Alfa)
which can consist the additional roots besides the roots of $(7).$
Consideraton of $(9)$ as the linear by $\{s, s^2\}$ leads to the solution
$$\begin{pmatrix} v\\ v^2\end{pmatrix} =
\dfrac1{\Delta}\begin{pmatrix} \Delta_1\\ \Delta_2\end{pmatrix},\tag{10}$$
where
\begin{align}
\Delta&=
\begin{vmatrix}
-6(27z^5+11z^4-4z^2+2)&9(11z^4-4z^2+2)\\[4pt]
\genfrac{.}{.}{0}{0}{2(847z^8-6534z^7-1012z^6+2808z^5}{+564z^4-1620z^3-184z^2+28)}
&\genfrac{.}{.}{0}{0}{3(2079z^7+484z^6}{-1188z^5-264z^4+810z^3+120z^2-16)}
\end{vmatrix}\\[4pt]
&=-18(65450z^{12}-35937z^{11}-41272z^{10}+28512z^9+28976z^8-21060z^7\\
&-2960z^6+6048z^5+988z^4-1620z^3-176z^2+24),\\
\end{align}
$$\Delta = -18d(z),\tag{11}$$
$$\begin{align}
d&=65450z^{12}-35937z^{11}-41272z^{10}+28512z^9+28976z^8-21060z^7\\
&-2960z^6+6048z^5+988z^4-1620z^3-176z^2+24,
\end{align}\tag{12}$$
\begin{align}
\Delta_1&=
\begin{vmatrix}
11z^6-108z^5-15z^4+6z^2+2&9(11z^4-4z^2+2)\\[4pt]
\genfrac{.}{.}{0}{0}{9(1617z^9+242z^8-1743z^7-308z^6}{+714z^5+168z^4-210z^3-56z^2+8)}
&\genfrac{.}{.}{0}{0}{3(2079z^7+484z^6-1188z^5}{-264z^4+810z^3+120z^2-16)}
\end{vmatrix}\\[4pt]
&=-6(228690z^{13}+145541z^{12}-297891z^{11}-117876z^{10}+209952z^9+87762z^8\\
&-102762z^7-24408z^6+28512z^5+8532z^4-6480z^3-2016z^2+232),\\
\end{align}
$$\Delta_1=-6d_1(z),\tag{13}$$
$$\begin{align}
d_1(z)&=228690z^{13}+145541z^{12}-297891z^{11}-117876z^{10}+209952z^9+87762z^8\\
&-102762z^7-24408z^6+28512z^5+8532z^4-6480z^3-2016z^2+232,
\end{align}\tag{14}$$
\begin{align}
\Delta_2&=
\begin{vmatrix}
-6(27z^5+11z^4-4z^2+2)&11z^6-108z^5-15z^4+6z^2+2\\[4pt]
\genfrac{.}{.}{0}{0}{2(847z^8-6534z^7-1012z^6+2808z^5}{+564z^4-1620z^3-184z^2+28)}
&\genfrac{.}{.}{0}{0}{9(1617z^9+242z^8-1743z^7-308z^6}{+714z^5+168z^4-210z^3-56z^2+8)}
\end{vmatrix},\\[4pt]
&=-2(1188110z^{14}+493317z^{13}-516938z^{12}-678645z^{11}+126096z^{10}+450036z^9\\
&+103236z^8-226476z^7-46980z^6+59940z^5+17100z^4-14580z^3-4088z^2+488).\\
\end{align}
$$\Delta_2 = -2d_2(z),\tag{15}$$
$$\begin{align}
d_2(z)&=1188110z^{14}+493317z^{13}-516938z^{12}-678645z^{11}+126096z^{10}+450036z^9\\
&+103236z^8-226476z^7-46980z^6+59940z^5+17100z^4-14580z^3-4088z^2+488,\\
\end{align}\tag{16}$$
$(10),(12),(14),(16)$ allow to write
$$v=\dfrac{d_1(z)}{3d(z)},\quad d_1(z)^2 = d_2(z)d(z),$$
or
$$
\begin{cases}
76977054000z^{25}-25462683400z^{26}-14469601850z^{24}-128299931694z^{23}+\\
47986668444z^{22}+144016029432z^{21}-67667920532z^{20}-106475101020z^{19}+\\
50533574480z^{18}+61655670720z^{17}-26340071868z^{16}-27611975304z^{15}+\\
9481920960z^{14}+10104266592z^{13}-2350471584z^{12}-3006244368z^{11}+331237872z^{10}+\\721643904z^9+10846384z^8-136453248z^7-19240960z^6+\\
17418240z^5+5038080z^4-1140480z^3-563776z^2+30976=0\\
v=\dfrac{d_1(z)}{3d(z)},
v>1,\quad 2 \ge z\ge1.
\end{cases}\tag{17}$$
The first equation allows partial decomposition to the form of
$$\begin{align}
&(z-1)(11z^4-4z^2+2)\times\\
&(1157394700z^{21}-2341562300z^{20}-1262982325z^{19}+3296484752z^{18} +1297048050z^{17}\\
&-2954973078z^{16}-802331850z^{15}+1662261954z^{14}+511593870z^{13}-621625938z^{12}\\
&-234164016z^{11}+160737876z^{10}+79849092z^9-29795100z^8-22817208z^7+2159304z^6\\
&+4347504z^5+563184z^4-402832z^3-117712z^2+7744z+7744),
\end{align}\tag{18}$$
where the last polynomial factor has the roots
$$\begin{align}
&z\in\{-0.65254852+0.411596443i,\quad
-0.65254852-0.411596443i,\quad
-0.533266085,\\
&-0.463163014+0.296052158i,\quad
-0.463163014-0.296052158i,\\
&-0.423458642-0.209546484i,\quad
-0.423458642+0.209546484i,\\
&-0.242560904-0.338675996i,\quad
-0.242560904+0.338675996i,\\
&-0.220398679-0.318470135i,\quad
-0.220398679+0.318470135i,\\
&0.334079252+0.005204691i,\quad
0.334079252-0.005204691i,\\
&0.397903595+0.409264586i,\quad
0.397903595-0.409264586i,\\
&0.553362795-0.024077867i,\quad
0.553362795+0.024077867i,\\
&0.658848407+0.496670038i,\quad
0.658848407-0.496670038i,\\
&0.883475687,\quad 1.788793855\}.\\
\end{align}$$
Easy to see that this leads to the result $(8).$
Best Answer
maybe this following solution is by Vasc?
since use Cauchy-Schwarz inequality,we have $$(a^3+3b)(a+3b)\ge (a^2+3b)^2$$ It suffices to show that $$\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\ge 6$$ By Holder $$\left(\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\right)^2[\sum_{cyc}(a^2+3b)(a+3b)]\ge[\sum_{cyc}(a^2+3b)]^3=[a^2+b^2+c^2+9]^3$$ it is enought to show that $$\left(a^2+b^2+c^2+9\right)^3\ge 36\sum_{cyc}(a^2+3b)(a+3b)$$ let $p=a+b+c=3,q=ab+bc+ac\le 3$,we have $$\sum_{cyc}(a^2+3b)(a+3b)=108-24q+3[abc+\sum_{cyc}a^2b]$$ use this well know:see inequality $$abc+\sum_{cyc}a^2b\le 4$$ we get $$\sum_{cyc}(a^2+3b)(a+3b)\le 24(5-q)$$ and $$a^2+b^2+c^2+9=2(9-q)$$ It suffices to show that $$(9-q)^3\ge 108(5-q)$$ which is true,because equivalent $$(q-3)^2(21-q)\ge 0$$