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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

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The idea is same as that of the solution of this question but is a little more complex in this case.


The original problem is equivalent to finding the minimum value of $$ \max (\frac{3}{3-2c}, \frac{3a}{3-2d}, \frac{3b}{3-2e}) \cdot \frac{9 - c - 2d - 3e}{2+3a +4b} $$ in the region $\{(a, b, c, d, e) \mid 0 \leq b \leq a \leq 1 \wedge 0 \leq c \leq d \leq e \leq 1\}$. Consider three cases listed below.

  • Case 1: $\frac{3a}{3-2d} \geq \frac{3}{3-2c} \ \wedge \ \frac{3a}{3-2d} \geq \frac{3b}{3-2e}$; $\ $that is, $a \geq \frac{3-2d}{3-2c}\ \wedge\ b \leq \frac{3-2e}{3-2d}\cdot a$;

  • Case 2: $\frac{3b}{3-2e} \geq \frac{3}{3-2c}\ \wedge\ \frac{3b}{3-2e} \geq \frac{3a}{3-2d}$; $\ $that is, $b \geq \frac{3-2e}{3-2c}\ \wedge\ a \leq \frac{3-2d}{3-2e} \cdot b$;

  • Case 3: $\frac{3}{3-2c} \geq \frac{3a}{3-2d}\ \wedge\ \frac{3}{3-2c} \geq \frac{3b}{3-2e}$; $\ $that is, $a \leq \frac{3-2d}{3-2c}\ \wedge\ b \leq \frac{3-2e}{3-2c}$.


Case 1: The problem can be stated as \begin{align} \text{minimize}\quad & f(a, b, c, d, e) = \frac{3a}{2 + 3a + 4b}\cdot \frac{9 - c - 2d - 3e}{3 - 2d} \\ \text{subject to}\quad & a \geq \frac{3-2d}{3-2c}\ \wedge\ b \leq \frac{3-2e}{3-2d}\cdot a \quad\quad (\text{by the condition of Case 1}) \\ & 0 \leq b \leq a \leq 1 \\ & 0 \leq c \leq d \leq e \leq 1 \end{align}

For fixed $a, c, d, e$, to minimize $f$, $b$ should be maximized, i.e., $b = \frac{3 - 2e}{3-2d}\cdot a \leq a$. Thus $f$ can rewritten as $$ f(a, c, d, e) = \frac{3a}{2 + 3a + 4a\cdot \frac{3-2e}{3-2d}}\cdot \frac{9 - c - 2d - 3e}{3 - 2d} $$ For fixed $c, d, e$, to minimize $f$, $a$ should be minimized, i.e., $a = \frac{3-2d}{3-2c} \leq 1$. $f$ then can be written as (after some transformation) $$ f(c, d, e) = \frac{27 - 3c - 6d - 9e}{27 - 4c - 6d - 8e} = \frac{3}{4} + \frac{1}{4} \cdot \frac{27 - 6d - 12e}{27 - 4c - 6d - 8e} $$

For fixed $d, e$, to minimize $f$, $c$ should equal $0$. Hence $f$ can be rewritten as $$ f(d, e) = \frac{3}{4} + \frac{1}{4} \cdot \frac{27 - 6d - 12e}{27 - 6d - 8e} = \frac{3}{4} + \frac{1}{4} \cdot (1 - \frac{4e}{27 - 6d - 8e}) $$ For fixed $e$, to minimize $f$, $d$ should equal $e$. Finally, we write $f$ as $$ f(e) = \frac{3}{4} + \frac{1}{4} \cdot (1 - \frac{4e}{27 - 14e}) $$ and it is minimized when $e = 1$ and the corresponding value is $\frac{12}{13}$.


Case 2: The problem can be stated as \begin{align} \text{minimize}\quad & g(a, b, c, d, e) = \frac{3b}{2 + 3a + 4b}\cdot \frac{9 - c - 2d - 3e}{3 - 2e} \\ \text{subject to}\quad & b \geq \frac{3-2e}{3-2c}\ \wedge\ a \leq \frac{3-2d}{3-2e} \cdot b \quad\quad (\text{by the condition of Case 2}) \\ & 0 \leq b \leq a \leq 1 \\ & 0 \leq c \leq d \leq e \leq 1 \end{align}

Similar to Case 1, we set $b = \frac{3-2e}{3-2c}$ and $a = \frac{3-2d}{3-2e}\cdot b$ and $g$ can be rewritten as $$ g(c, d, e) = \frac{27 - 3c - 6d - 9e}{27 - 4c - 6d - 8e} $$ same as $f(c, d, e)$ in Case 1. So the minimum value is still $\frac{12}{13}$.


Case 3: The problem can be stated as \begin{align} \text{minimize}\quad & h(a, b, c, d, e) = \frac{3}{2 + 3a + 4b}\cdot \frac{9 - c - 2d - 3e}{3 - 2c} \\ \text{subject to}\quad & a \leq \frac{3-2d}{3-2c}\ \wedge\ b \leq \frac{3-2e}{3-2c} \quad\quad (\text{by the condition of Case 3}) \\ & 0 \leq b \leq a \leq 1 \\ & 0 \leq c \leq d \leq e \leq 1 \end{align}

For fixed $c, d, e$, $h$ is minimized when $a$ and $b$ are maximized, i.e., $a = \frac{3-2d}{3-2c}$ and $b = \frac{3-2e}{3-2c}$. Then $h$ becomes $$ h(c, d, e) = \frac{27 - 3c - 6d - 9e}{27 - 4c - 6d - 8e} $$ same as $f(c, d, e)$. Thus the minimum value is $\frac{12}{13}$.


Summary: the greatest $k$ is $\frac{12}{13}$.

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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 $0\leq c\leq d\leq e\leq 1$.

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  • $\begingroup$ Thanks. In fact, a better bound of $k=1/3$ follows from your first inequality, so the rest is not needed. The natural question would then be to ask for the best such constant $k$. I've edited the question. $\endgroup$ – pi66 Jul 13 '16 at 9:47
  • $\begingroup$ @pi66: Ah, you are right. I've misunderstood something. $\endgroup$ – mathlove Jul 13 '16 at 10:28

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