Let
$$\sqrt{3a+b+c} = x, \sqrt{3b+c+d} = y, \sqrt{3c+d+a} = z, \sqrt{3d+a+b} = w,$$
then we have
$$x^2 = 3a + b + c, \quad y^2 = 3b + c + d,\\ z^2 = 3c + d + a, \quad w^2 = 3d + a + b.$$
Therefore, we have
$$\begin{aligned} 15a &= 5(3a + b + c) - 2(3b + c + d) - (3c + d + a) + (3d + a + b) \\ &= 5x^2 - 2y^2 - z^2 + w^2. \end{aligned}$$
Hence we can conclude that
$$a = \frac{5x^2 - 2y^2 - z^2 + w^2}{15},$$
and similarly we have
$$b = \frac{5y^2 - 2z^2 - w^2 + x^2}{15},\\ c = \frac{5z^2 - 2w^2 - x^2 + y^2}{15},\\ d = \frac{5w^2 - 2x^2 - y^2 + z^2}{15},$$
all non negative as well. Now since
$$a + b + c + d = 4,$$
we have
$$\begin{aligned} x^2 + y^2 + z^2 + w^2 &= (3a + b + c) + (3b + c + d) \\ &\quad + (3c + d + a) + (3d + a + b) \\ &= 20 \end{aligned}$$
and we need to prove that
$$\begin{aligned} \sum_{cyc} x \left( \frac{5x^2 - 2y^2 - z^2 + w^2}{15} \right) &\geq 4\sqrt{5} \\ \Leftrightarrow \sum_{cyc} (5x^3 - 2xy^2 - xz^2 + xw^2) &\geq 60\sqrt{5}. \end{aligned}$$
Now by AM-GM we know that
$$\sum_{cyc} (x^3 - 2xy^2 + xw^2) = \sum_{cyc} (x^3 - 2xy^2 + yx^2) \geq 0$$
and again by AM-GM
$$\sum_{cyc} (x^3 - xz^2) = \frac{1}{3} \sum_{cyc} (x^3 + z^3 + z^3 - 3xz^2) \geq 0.$$
Hence we only need to prove that
$$\sum_{cyc} 3x^3 \geq 60\sqrt{5}$$
which is again AM-GM:
$$\begin{aligned} \sum_{cyc} (3x^3 + 3x^3 + 15\sqrt{5}) &\geq \sum_{cyc} 3\sqrt[3]{3 \times 3 \times 15\sqrt{5}x^6} \\ &= \sum_{cyc} 9\sqrt{5}x^2 = 180\sqrt{5} \\ \Leftrightarrow \sum_{cyc} 3x^3 &\geq \frac{180 - 60}{2}\sqrt{5} = 60\sqrt{5} \end{aligned}$$
as desired.

$$x^{2} = y^{2} = z^{2} = w^{2} = 5$$
which is when
$$a = b = c = d = 1.$$

Another Solution: By generalized Cauchy inequality,
$$\left( \sum_{cyc} a\sqrt{3a+b+c} \right)^2 \left( \sum_{cyc} \frac{a}{3a+b+c} \right) \geq (a+b+c+d)^3 = 64.$$
Therefore it suffices to show that
$$\frac{a}{3a+b+c} + \frac{b}{3b+c+d} + \frac{c}{3c+d+a} + \frac{d}{3d+a+b} \leq \frac{4}{5}.$$
By subtracting the above inequality multiplied by 3 from 4, we get that it is equivalent to
$$\frac{b+c}{3a+b+c} + \frac{c+d}{3b+c+d} + \frac{d+a}{3c+d+a} + \frac{a+b}{3d+a+b} \geq \frac{8}{5}.$$
By Cauchy inequality,
$$(L.H.S.) \left( \sum_{cyc} (b+c)(3a+b+c) \right) \geq 4(a+b+c+d)^2.$$
Therefore it suffices to show that
$$\sum_{cyc} (b+c)(3a+b+c) = 2 \left( \sum_{cyc} a^2 \right) + 5 \left( \sum_{cyc} ab \right) + 6ac + 6bd \leq \frac{5}{2} (a+b+c+d)^2,$$
which becomes, after expansion and simplification,
$$ac + bd \leq \frac{1}{2}(a^2 + b^2 + c^2 + d^2).$$
The last inequality is the sum of two AM-GMs, and so we're done.