maximize $\log(4)c+\log(3)a+\log(2)x$ if $a+c+x+y=1$, $(a+c)^2+(x+y)^2+2xc\leq 1-2\gamma$, $0\leq \gamma \leq 1/4$

Question

Edit: I am using the natural logarithm in what follows.

I want to figure out how to show by hand that the maximum of $$\log(4)c+\log(3)a+\log(2)x$$ when $$a\geq 0, c\geq 0, x \geq 0, y \geq 0,$$ $$a+c+x+y=1,$$ $$(a+c)^2+(x+y)^2+2xc\leq 1-2\gamma,$$ where $\gamma$ is a fixed constant such that $4/25\leq \gamma \leq 1/4$, is $\frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)$, which is given by $a=\frac{1+\sqrt{1-4\gamma}}{2}$, $x=1-a=\frac{1-\sqrt{1-4\gamma}}{2}$, $c=y=0$.

I have tried using Lagrange multipliers by changing the last constraint to an equality, but it becomes very messy and I get stuck.

I would also be happy if I could simply show that $\log(4)c+\log(3)a+\log(2)x \leq \frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)$ for all such $a,c,x,y$. A method I have attempted is to use the concavity of $\log(x)$. The equation of the line that passes through $(2, \log(2))$ and $(3, \log(3))$ is $L(x)=\log(3/2)x-\log(9/8)$. Then by concavity of $\log(x)$ we have $\log(x)\leq L(x)$ for all integers $x$. Then

\begin{align*}&\log(4)c+\log(3)a+\log(2)x\\ &\leq L(4)c+L(3)a+L(2)x\\ &=\log(3/2) (4c+3a+2x+y)-\log(9/8)\\ &=\log(3/2)\left(\frac{5+3}{2}c+\frac{5+1}{2}a+\frac{5-1}{2}x+\frac{5-3}{2}y\right)-\log(9/8)\\ &=\log(3/2)\left(\frac{5+\sqrt{1-4\gamma}}{2}\right)-\log(9/8)+\log(3/2)\left(\frac{a-x+3(c-y)-\sqrt{1-4\gamma}}{2}\right)\\ &=\frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)+\log(3/2)\left(\frac{a-x+3(c-y)-\sqrt{1-4\gamma}}{2}\right).\\ \end{align*} That means that to obtain the inequality I want, it would suffice to show that $a-x\leq 3(y-c)+\sqrt{1-4\gamma}$. However, I have not had success in proving this inequality. Any suggestions or help is immensely appreciated.

Another option would be to use perturbation techniques, but I do not have experience in this. Thank you for your help.

Answer 1

$\frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)$ is not the maximum of $$f(c,a,x):=\log(4)c+\log(3)a+\log(2)x$$

For example, for $\gamma=4/25$, we have $$f(4/5,0,0)=\frac{\log(256)}{5}\color{red}{\gt}\frac{\log(162)}{5}=\frac{\log(6)}{2}+\frac{\sqrt{1-4(4/25)}}{2}\log(3/2)$$

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In the following, let us prove that the maximum of $\log(4)c+\log(3)a+\log(2)x$ is $$\begin{cases}\log(2)(1+\sqrt{1-4\gamma})&\text{if $\ 4/25\le \gamma\lt\frac{\log(2)\log(4/3)}{(\log(8/3))^2}$} \\\\\frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)&\text{if $\ \frac{\log(2)\log(4/3)}{(\log(8/3))^2}\le\gamma\le 1/4$}\end{cases}$$

We want to find the maximum of $f(c,a,x)$ under the condition that $$a\geq 0, c\geq 0, x \geq 0, 1-a-c-x \geq 0,$$ $$(a+c)^2+(1-a-c)^2+2xc\leq 1-2\gamma,4/25\leq \gamma \leq 1/4$$

Case 1 : $c=0$

We want to find the maximum of $f(0,a,x)$ under the condition that $$0\le x\le 1-a,\frac{1-\sqrt{1-4\gamma}}{2}\le a\le\frac{1+\sqrt{1-4\gamma}}{2}, 4/25\leq \gamma \leq 1/4$$

Therefore, we get $$\begin{align}f(0,a,x)&\le f(0,a,1-a) \\\\&=\log(3/2)a+\log(2) \\\\&\le\log(3/2)\frac{1+\sqrt{1-4\gamma}}{2}+\log(2) \\\\&=\frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)\end{align}$$

Case 2 : $c\gt 0$

We want to find the maximum of $f(c,a,x)$ under the condition that $$a\ge 0, c\gt 0, 0\le 1-a-c,$$ $$0\le x\leq \min\bigg(1-a-c,\frac{1-2\gamma-(a+c)^2-(1-a-c)^2}{2c}\bigg),$$ $$0\le \frac{1-2\gamma-(a+c)^2-(1-a-c)^2}{2c}, 4/25\leq \gamma \leq 1/4$$ Note here that $$\min\bigg(1-a-c,\frac{1-2\gamma-(a+c)^2-(1-a-c)^2}{2c}\bigg)=\begin{cases}1-a-c&\text{if $\ a(1-a-c)\ge \gamma$} \\\\\frac{1-2\gamma-(a+c)^2-(1-a-c)^2}{2c}&\text{if $\ a(1-a-c)\lt \gamma$}\end{cases}$$

Case 2-1 : $a(1-a-c)\ge \gamma$

We have $0\le x\leq 1-a-c$ from which we obtain $$f(c,a,x)\le f(c,a,1-a-c)=\log(2)c+\log(3/2)a+\log(2):=g(c,a)$$

Suppose that $a=0$. Then, there is no $\gamma$ such that $0\ge\gamma$ and $4/25\leq \gamma \leq 1/4$. So, $a\gt 0$.

So, we want to find the maximum of $g(c,a)$ under the condition that $$0\lt c\le 1-a-\frac{\gamma}{a},\frac{1-\sqrt{1-4\gamma}}{2}\lt a\lt\frac{1+\sqrt{1-4\gamma}}{2}, 4/25\leq \gamma \leq 1/4$$ so $$g(c,a)\le g\bigg(1-a-\frac{\gamma}{a},a\bigg)=-\log(2)\frac{\gamma}{a}-\log(4/3)a+\log(4):=h(a)$$ We want to find the maximum of $h(a)$ under the condition that $$\frac{1-\sqrt{1-4\gamma}}{2}\lt a\lt\frac{1+\sqrt{1-4\gamma}}{2}, 4/25\leq \gamma \leq 1/4$$

We see that $h'(a)$ is decreasing with $$h'(a)=0\iff a=\sqrt{\frac{\log(2)}{\log(4/3)}\gamma}$$

If $4/25\le \gamma\lt \frac{\log(2)\log(4/3)}{(\log(8/3))^2}$, then $$\frac{1-\sqrt{1-4\gamma}}{2}\lt \sqrt{\frac{\log(2)}{\log(4/3)}\gamma}\lt \frac{1+\sqrt{1-4\gamma}}{2}$$ from which we have $$h(a)\le h\bigg(\sqrt{\frac{\log(2)}{\log(4/3)}\gamma}\bigg)=\log(4)-2\sqrt{\log(2)\log(4/3)\gamma}$$

If $\frac{\log(2)\log(4/3)}{(\log(8/3))^2}\le\gamma\le\frac 14$, then $$\frac{1+\sqrt{1-4\gamma}}{2}\le \sqrt{\frac{\log(2)}{\log(4/3)}\gamma}$$ So, $h'(a)$ is positive in $\frac{1-\sqrt{1-4\gamma}}{2}\lt a\lt\frac{1+\sqrt{1-4\gamma}}{2}$, so $h(a)$ is increasing from which we have $$h(a)\lt h\bigg(\frac{1+\sqrt{1-4\gamma}}{2}\bigg)=\frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)$$

Case 2-2 : $a(1-a-c)\lt \gamma$

We have $$0\le x\leq \frac{1-2\gamma-(a+c)^2-(1-a-c)^2}{2c}$$ from which we obtain $$\begin{align}f(c,a,x)&\le f\bigg(c,a,\frac{1-2\gamma-(a+c)^2-(1-a-c)^2}{2c}\bigg) \\\\&=\log(2)c+\log(3/4)a+\log(2)\frac{-\gamma-a^2+a+c}{c}:=j(c,a)\end{align}$$ where $$\frac{\partial j}{\partial c}=\log(2)\frac{c^2+\gamma+a^2-a}{c^2}$$

We want to maximize $j(c,a)$ under the condition that $$0\le a\lt \frac{1+\sqrt{1-4\gamma}}{2}, 0\lt c\le 1-a,\frac{a-a^2-\gamma}{a}\lt c$$ $$\frac{1-2a-\sqrt{1-4\gamma}}{2}\le c\le \frac{1-2a+\sqrt{1-4\gamma}}{2}, 4/25\leq \gamma \leq 1/4$$

Case 2-2-1 : $\frac{1-\sqrt{1-4\gamma}}{2}\lt a\lt\frac{1+\sqrt{1-4\gamma}}{2}$

Then, we have

$$\sqrt{a-a^2-\gamma}\ge\frac{a-a^2-\gamma}{a}$$

$$\small\frac{1-2a-\sqrt{1-4\gamma}}{2}\le 0\le \frac{a-a^2-\gamma}{a}\lt \frac{1-2a+\sqrt{1-4\gamma}}{2}\lt 1-a$$

So, we want to maximize $j(c,a)$ under the condition that $$\frac{1-\sqrt{1-4\gamma}}{2}\lt a\le\frac{1+\sqrt{1-4\gamma}}{2},\frac{a-a^2-\gamma}{a}\lt c\le \frac{1-2a+\sqrt{1-4\gamma}}{2}, 4/25\leq \gamma \leq 1/4$$

Using that $$\frac{\partial j}{\partial c}=\log(2)\frac{c^2+\gamma+a^2-a}{c^2}\ge 0\iff c\ge \sqrt{a-a^2-\gamma}$$ we have $$\begin{align}j(c,a)&\le\max\bigg(j\bigg(\frac{a-a^2-\gamma}{a},a\bigg),j\bigg(\frac{1-2a+\sqrt{1-4\gamma}}{2},a\bigg)\bigg) \\\\&=\begin{cases}j\bigg(\frac{a-a^2-\gamma}{a},a\bigg)&\text{if $\ a\ge\frac{1+\sqrt{1-4\gamma}}{4}$} \\\\j\bigg(\frac{1-2a+\sqrt{1-4\gamma}}{2},a\bigg)&\text{if $\ a\lt\frac{1+\sqrt{1-4\gamma}}{4}$}\end{cases}\end{align}$$

Case 2-2-1-1 : $\frac{1+\sqrt{1-4\gamma}}{4}\le\frac{1-\sqrt{1-4\gamma}}{2}\ (\le a)$ which holds only when $2/9\le\gamma\le 1/4$

$$\begin{align}j(c,a)&\le j\bigg(\frac{a-a^2-\gamma}{a},a\bigg) \\\\&=h(a) \\\\&\lt h\bigg(\frac{1+\sqrt{1-4\gamma}}{2}\bigg) \\\\&=\frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)\end{align}$$

Case 2-2-1-2 : $\frac{1-\sqrt{1-4\gamma}}{2}\lt\frac{1+\sqrt{1-4\gamma}}{4}\lt\frac{1+\sqrt{1-4\gamma}}{2}$ which holds only when $4/25\le\gamma\lt 2/9$

Case 2-2-1-2-1 : $\frac{1-\sqrt{1-4\gamma}}{2}\lt a\le\frac{1+\sqrt{1-4\gamma}}{4}$

$$\begin{align}j(c,a)&\le j\bigg(\frac{1-2a+\sqrt{1-4\gamma}}{2},a\bigg) \\\\&=\log(2)(1+\sqrt{1-4\gamma})-\log(4/3)a \\\\&\lt \log(2)(1+\sqrt{1-4\gamma})-\log(4/3)\frac{1-\sqrt{1-4\gamma}}{2} \\\\&=\frac{\log(3)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(16/3)\end{align}$$

Case 2-2-1-2-2 : $\frac{1+\sqrt{1-4\gamma}}{4}\lt a\lt\frac{1+\sqrt{1-4\gamma}}{2}$

$$\begin{align}j(c,a)&\le j\bigg(\frac{1-2a+\sqrt{1-4\gamma}}{2},a\bigg) \\\\&=\log(2)(1+\sqrt{1-4\gamma})-\log(4/3)a \\\\&\lt\log(2)(1+\sqrt{1-4\gamma})-\log(4/3)\frac{1+\sqrt{1-4\gamma}}{4} \\\\&=\frac{\log(12)}{4}+\frac{\sqrt{1-4\gamma}}{4}\log(12)\end{align}$$

Case 2-2-2 : $0\le a\le \frac{1-\sqrt{1-4\gamma}}{2}$

Since $\frac{\partial j}{\partial c}=\log(2)\frac{c^2+\gamma+a^2-a}{c^2}\gt 0$, we see that $j(c,a)$ is increasing on $c$. We have

$$\small\frac{a-a^2-\gamma}{a}\le 0\le \frac{1-2a-\sqrt{1-4\gamma}}{2}\le \frac{1-2a+\sqrt{1-4\gamma}}{2}\le 1-a$$

So, we want to maximize $j(c,a)$ under the condition that $$0\le a\le \frac{1-\sqrt{1-4\gamma}}{2}, \frac{1-2a-\sqrt{1-4\gamma}}{2}\le c\le \frac{1-2a+\sqrt{1-4\gamma}}{2}, 4/25\leq \gamma \leq 1/4$$

Hence, we get $$\begin{align}j(c,a)&\le j\bigg(\frac{1-2a+\sqrt{1-4\gamma}}{2},a\bigg) \\\\&=\log(2)(1+\sqrt{1-4\gamma})-\log(4/3)a \\\\&\le\log(2)(1+\sqrt{1-4\gamma})\end{align}$$

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Conclusion : The maximum of $\log(4)c+\log(3)a+\log(2)x$ is $$\begin{cases}\log(2)(1+\sqrt{1-4\gamma})&\text{if $\ 4/25\le \gamma\lt\frac{\log(2)\log(4/3)}{(\log(8/3))^2}$} \\\\\frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2)&\text{if $\ \frac{\log(2)\log(4/3)}{(\log(8/3))^2}\le\gamma\le 1/4$}\end{cases}$$

Answer 2

$\color{brown}{\textbf{Preliminary transformations.}}$

The task is to maximize the function $$f(x,a,c) = (x+2c)\log 2 + a\log 3\tag1$$ over non-negative $a,c,x,y,$ under the conditions $$x+y+a+c = 1,\quad (x+y)^2+(c+a)^2+2cx\le 1-2\gamma,\tag2$$ or \begin{cases} x+y+a+c = 1\\ (c+a)(x+y) - cx\ge\gamma\\ x\ge0,\quad y\ge0,\quad a\ge0,\quad c\ge0,\tag3 \end{cases} where $$\gamma\in\left[0,\dfrac14\right].\tag4$$

Denote $$b= a+c,\quad z = 1-y,\quad g(z,b,c) = f(z-b,b-c,c),$$ then $$g(z,b,c) = (z-b+2c)\log2+(b-c)\log3\tag5,$$

\begin{cases} b(1-b) - c(z-b)\ge\gamma\\ 0\le c\le b\le z\le1.\tag6 \end{cases}

$\color{brown}{\textbf{Searching of the global maximum.}}$

Maximum of the linear function $g(z,b,c)$ corresponds to the edges of area, wherein maximum of the linear function with the linear constraints - to the vertices.

Since from $(5)$ $$g(z,b,c) = z\log2+b\log\dfrac32+c\log\dfrac43,$$ then the global maximum corresponds to the point with the maximal distance from the start of the coordinates at the certain direction.

Vertices.

If c=0 then

$$b^2-b + \gamma = 0,$$ $$\max g_v(z,b,0) = g\left(1,\frac12+\frac{\sqrt{1-4\gamma}}2,0\right),$$ $$\max g_v(z,b,0) = \left(\frac12-\frac{\sqrt{1-4\gamma}}2\right)\log2 +\left(\frac12+\frac{\sqrt{1-4\gamma}}2\right)\log3.$$

If b=c then \begin{cases} b(1-z)=\gamma\\ z = b, \end{cases} $$\max g_v(z,b,b) = g\left(\frac12+\frac{\sqrt{1-4\gamma}}2, \frac12+\frac{\sqrt{1-4\gamma}}2,\frac12+\frac{\sqrt{1-4\gamma}}2\right),$$ $$\max g_v(z,b,b) = \left(1+\sqrt{1-4\gamma}\right)\log2.$$

If z=b then \begin{cases} b(1-b)=\gamma\\ c=b, \end{cases} $$\max g_v(b,b,c) = \left(1+\sqrt{1-4\gamma}\right)\log2.$$

If z=1 then \begin{cases} (b-c)(1-b)=\gamma\\ c = 0 \end{cases} $$\max g_v(1,b,c) = g\left(1,\frac12+\frac{\sqrt{1-4\gamma}}2,0\right),$$ $$\max g_v(1,b,c) = \left(\frac12-\frac{\sqrt{1-4\gamma}}2\right)\log2 +\left(\frac12+\frac{\sqrt{1-4\gamma}}2\right)\log3.$$

The greatest value over vertices is $$\color{brown}{\mathbf{g_v(z,b,c) = \begin{cases} \left(1+\sqrt{1-4\gamma}\right)\log2,\quad\text{if}\quad \gamma\in[0,\gamma_v)\\[4pt] \left(\frac12-\frac{\sqrt{1-4\gamma}}2\right)\log2 +\left(\frac12+\frac{\sqrt{1-4\gamma}}2\right)\log3, \quad\text{if}\quad \gamma\in[\gamma_v,0.25], \end{cases}}}\tag7$$ where $$\color{brown}{\mathbf{\gamma_v= \dfrac14 - \dfrac14\left(\dfrac{\log\,^3/_2}{\log\,^8/_3}\right)^2\approx0.20728.}}\tag8$$

Optimization task.

Optimization task for non-linear constraint $(6.1)$ can be solved by Lagrange multipliers method, applied to the function $$G(z,b,c,\lambda) = (z-b+2c)\log2+(b-c)\log3+\lambda(b-b^2+bc-cz - \gamma).$$ The stationary points of $G$ can be defined from the system $G'_z = G'_b = G'_c = G'_\lambda = 0,$ or \begin{cases} \log2-\lambda c = 0\\ -\log2+\log3+\lambda(1-2b+c) = 0\\ 2\log2-\log3+\lambda(b-z) = 0\\ b(1-b+c)-cz - \gamma = 0. \end{cases}

Then, taking in account $(6.1),$ $$ \begin{cases} \log2-\lambda c = 0\\ \log3+\lambda(1-2b) = 0\\ \log2 + \lambda(1-b+c-z) = 0\\ b(1-b+c)-cz - \gamma = 0 \end{cases}\Rightarrow \begin{cases} (2b-1)\log2= c\log3\\ z=1-b+2c\\ b(1-b+c) - c(1-b+2c) = \gamma, \end{cases}$$ \begin{cases} c=r(2b-1)\\ z=(4r-1)b-(2r-1)\\ b(1-b)+c(2b-1-2c) = \gamma\\ 0\le c\le b \le z \le1\\ r=\dfrac{\log2}{\log3} = \log_23\approx0.63093,\tag9 \end{cases}

\begin{cases} c=r(2b-1)\\ z=1-b+2c\\ b(1-b)+(1-2r)(2b-1)^2=\gamma\\ b\in\left[\dfrac12,\dfrac{2r}{4r-1}\right]\approx[0.5,0.82814]\\ \gamma\in\left[0,\dfrac{(2r-1)^2}{(4r-1)^2}\right]\approx[0,0.02953],\tag{10} \end{cases}

$$b(1-b)+r(1-2r)(4b^2-4b+1) = \gamma,$$ $$s(b^2-b) + 2r^2-r+\gamma = 0,\quad\text{where}\quad s=8r^2-4r+1\approx1.66086,\tag{11}$$ $$b= \dfrac12 +\dfrac12\sqrt{\dfrac{1-4\gamma}s},\quad c=r\sqrt{\dfrac{1-4\gamma}s},\quad z= \dfrac12 +\dfrac{4r-1}2\sqrt{\dfrac{1-4\gamma}s},$$ $$g_m(z,b,c) = (4r-1)\sqrt{\dfrac{1-4\gamma}s}\log2+\left(\dfrac12+\dfrac{1-2r}2\sqrt{\dfrac{1-4\gamma}s}\right)\log3\\ = \left(\dfrac12+\dfrac{8r^2-4r+1}2\sqrt{\dfrac{1-4\gamma}s}\right)\log3 = \left(\dfrac12+\dfrac{s}2\sqrt{\dfrac{1-4\gamma}s}\right)\log3,$$ $$g_m(z,b,c) \le \dfrac{1+\sqrt s}2 \log3 < {1+1.3}2 = 1.15\log3,$$ $$g_v(z,b,c) > r\left(1+\sqrt{1-4\cdot0.3}\right)\log3 > 0.63(1+0.9)\log3 > g_m(z,b,c),$$

$$\color{brown}{\mathbf{\max g(z,b,c)=g_v(z,b,c).}}\tag{12}$$

Thus, the global maximum of $f(x,a,c)$ under the given conditions is defined by the formulas $\color{brown}{\mathbf{(7)-(8)}}\ $ (see also Wolfram Alpha plot).