Constant such that $\max\left(\frac{3}{3-2c},\frac{3a}{3-2d},\frac{3b}{3-2e}\right)\geq k\cdot\frac{2+3a+4b}{9-c-2d-3e}$ What is the greatest constant $k>0$ such that
$$\max\left(\frac{3}{3-2c},\frac{3a}{3-2d},\frac{3b}{3-2e}\right)\geq k\cdot\frac{2+3a+4b}{9-c-2d-3e}$$
for any $0\leq b\leq a\leq 1$ and $0\leq c\leq d\leq e\leq 1$?
This is a three-term version of this question. 
Update: As shown below by mathlove, $k=1/3$ holds since the following is true:
$$\frac{3}{3-2c}\cdot\frac{9-c-2d-3e}{2+3a+4b}\geq\frac{1}{3}.$$
WolframAlpha cannot find a global minimum of the product
$$\max\left(\frac{3}{3-2c},\frac{3a}{3-2d},\frac{3b}{3-2e}\right)\cdot\frac{9-c-2d-3e}{2+3a+4b}$$
but I'm not sure why that would be. The WolframAlpha link follows.
http://www.wolframalpha.com/input/?i=find+minimum+of+max(3%2F(3-2c),(3a)%2F(3-2d),(3b)%2F(3-2e))*(9-c-2d-3e)%2F(2%2B3a%2B4b)+for+0%3C%3Db%3C%3Da%3C%3D1+and+0%3C%3Dc%3C%3Dd%3C%3De%3C%3D1
 A: Yes, $\color{red}{k=\frac 17}$ works. 
First, 
$$\frac{3}{3-2c}\cdot\frac{9-c-2d-3e}{2+3a+4b}\ge \frac{3}{3-2\cdot 0}\cdot\frac{9-1-2\cdot 1-3\cdot 1}{2+3\cdot 1+4\cdot 1}=\frac 13$$
Second, when 
$$\frac{3a}{3-2d}\ge \frac{3}{3-2c}\implies a\ge\frac{3-2d}{3-2c}\ge \frac{3-2\cdot 1}{3-2\cdot 0}=\frac 13$$
we have
$$\frac{3a}{3-2d}\cdot\frac{9-c-2d-3e}{2+3a+4b}\ge \frac{3a}{3-2\cdot 0}\cdot\frac{9-1-2\cdot 1-3\cdot 1}{2+3a+4\cdot 1}=1-\frac{2}{a+2}$$$$\ge 1-\frac{2}{\frac 13+2}=\frac 17$$
Third, when
$$\frac{3b}{3-2e}\ge\frac{3}{3-2c}\implies b\ge \frac{3-2e}{3-2c}\ge \frac{3-2\cdot 1}{3-2\cdot 0}=\frac 13$$
we have
$$\frac{3b}{3-2e}\cdot\frac{9-c-2d-3e}{2+3a+4b}\ge \frac{3b}{3-2\cdot 0}\cdot\frac{9-1-2\cdot 1-3\cdot 1}{2+3\cdot 1+4\cdot b}=\frac 34-\frac{15}{4(4b+5)}$$$$\ge \frac 34-\frac{15}{4(4\cdot \frac 13+5)}=\frac{3}{19}$$
It follows from these that
$$\max\left(\frac{3}{3-2c},\frac{3a}{3-2d},\frac{3b}{3-2e}\right)\geq \color{red}{\frac 17}\cdot\frac{2+3a+4b}{9-c-2d-3e}$$
holds for any $0\leq b\leq a\leq 1$ and $0\leq c\leq d\leq e\leq 1$.
