Constant such that $\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$
The original problem is equivalent to finding the minimum value of $$ \max (\frac{5}{5-3c}, \frac{5b}{5-3d}) \cdot \frac{5-c-2d}{2+3b} $$ in the region $\{(b, c, d) \mid 0 \leq b \leq 1 \wedge 0 \leq c \leq d \leq 1\}$. We consider two cases below, namely,
Case 1: $\frac{5}{5-3c} \leq \frac{5b}{5-3d}$;
Case 2: $\frac{5}{5-3c} \geq \frac{5b}{5-3d}$.
Case 1: The problem can be stated as \begin{align} \text{minimize}\quad & f(b, c, d) = \frac{5-c-2d}{5-3d}\cdot \frac{5b}{2+3b} \\ \text{subject to}\quad & b \geq \frac{5-3d}{5-3c} \quad\quad (\text{by the condition of Case 1}) \\ & 0 \leq b \leq 1 \\ & 0 \leq c \leq d \leq 1 \end{align} Note that the term $\frac{5b}{2+3b} = \frac{5}{3} - \frac{10}{6+9b}$ in $f(b, c, d)$ is a strictly increasing function of $b$. To minimize $f(b, c, d)$, $b$ should be minimized, for fixed $c$ and $d$. Therefore, $$b = \frac{5-3d}{5-3c} \tag{1}$$ Substituting $b$ with (1) in $f$, we obtain that $$ f(b, c, d) = \frac{25-5c-10d}{25-6c-9d} = \frac{5}{6} + \frac{\frac{25}{6} - \frac{15}{6}d}{25 - 6c - 9d} $$ For fixed $d$, $f(b, c, d)$ is minimized when $c$ is minimized, i.e., $c = 0$. We further obtain that $$ f(b, c, d) = \frac{5}{6} + \frac{\frac{25}{6} - \frac{15}{6}d}{25 - 9d} = \frac{5}{6} + \frac{5}{18} - \frac{\frac{50}{18}}{25-9d} $$ Setting $d = 1$, we obtain the minimum value $\frac{5}{6} + \frac{5}{18} - \frac{50}{18\cdot16} = \frac{15}{16}$.
Case 2: The problem can be stated as: \begin{align} \text{minimize}\quad & g(b, c, d) = \frac{5-c-2d}{5-3c}\cdot \frac{5}{2+3b} \\ \text{subject to}\quad & b \leq \frac{5-3d}{5-3c} \quad\quad (\text{by the condition of Case 2}) \\ & 0 \leq b \leq 1 \\ & 0 \leq c \leq d \leq 1 \end{align} and the solution is similar to that of Case 1. First, the term $\frac{5}{2+3b}$ is a strictly decreasing function of $b$. Thus, we set $b = \frac{5-3d}{5-3c}$ and obtain $$ g(b, c, d) = \frac{25-5c-10d}{25-6c-9d} $$ which is the same function as that in Case 1. So the minimum value is still $\frac{15}{16}$.