It's easy to check, when $a = 1$, $b = 2$, $c = 3$ equality holds, hence by the Cauchy's inequality we have
$$\begin{aligned} & \frac{1}{4} \left( 2 \sqrt{\frac{3b}{a+b} \cdot \frac{4c}{a+b+c}} + 4 \sqrt[4]{\frac{3a}{a+b} \left( \frac{2(a+b)}{a+b+c} \right)^2} \cdot 1 \right) \le \\ & \le \frac{1}{4} \left( \frac{3b}{a+b} + \frac{4c}{a+b+c} + \frac{3a}{a+b} + \frac{2(a+b)}{a+b+c} + \frac{2(a+b)}{a+b+c} + 1 \right) = \\ & = \frac{1}{4}(3 + 4 + 1) = 2. \end{aligned}$$
Here equality holds for if and only if $\frac{3b}{a+b} = \frac{4c}{a+b+c}$ same as $\frac{3a}{a+b} = \frac{2(a+b)}{a+b+c} = 1$. Thus equality holds for $a > 0$, $b = 2a$, $c = 3a$.