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I am trying to maximize the function $$f(x,y,z)=\log(2)+\log(3/2)x+\log(2)y+\log(5/2)z$$ with the following constraints: $$x\geq 0, y\geq 0, z \geq 0,$$ $$x+y+z\leq 1,$$ $$x+y\geq 4/5,$$ $$(y+z)^2+2x-x^2-2xy\leq 1-2\gamma,$$ where $$0.24 \leq \gamma \leq 0.25.$$

I claim that the maximum value is $\frac{\log(12)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log\left( \frac{4}{3} \right)$, and that this maximum is obtained when $x=\frac{1+\sqrt{1-4\gamma}}{2}$, $y=\frac{1-\sqrt{1-4\gamma}}{2}$, and $z=0$.

I am trying to avoid using Lagrange Multipliers because it becomes complicated. I am wondering if there is another way. I would also be satisfied if I could show that $f(x,y,z)\leq \frac{\log(12)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log\left( \frac{4}{3} \right)$. Programs like Maple and Mathematica give me solutions for specific $\gamma$, but I would like to find a step by step way to show this for ANY $\gamma$. Thank you.

Note: I want to point out that we treat $\gamma$ as a FIXED constant that lies in the real interval $[0.24, 0.25]$. Also, all logarithms considered are real.

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  • $\begingroup$ How did you find the answer you claim to be true? $\endgroup$
    – skr
    May 7, 2020 at 3:34
  • $\begingroup$ I used Maple for $\gamma=0.24, 0.241, 0.245, 0.249, 0.2499, 0.25$. Of course this doesn't cover the entire real interval $[0.24, 0.25]$, but I've worked on similar maximization problems and the max value was similar. I haven't been able to apply the same methods here, though. $\endgroup$
    – Steve
    May 7, 2020 at 13:28
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    $\begingroup$ @Steve As I could test numerically, your claim is correct. $\endgroup$
    – Cesareo
    May 7, 2020 at 17:39
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    $\begingroup$ Maybe a bit easier is to solve the following equivalent problem. Let’s introduce a new variable $t=1-x-y-z$. Then we can exclude the variable $z$ from the problem, transforming it to a form Minimize the function $$g(x,y,z)=\log(5/3)x+\log(5/4)y+\log(5/2)t$$ with the following constraints: $$x\geq 0, \,y\geq 0,\, t \geq 0,$$ $$x+y+t\leq 1,$$ $$x+y\geq 4/5,$$ $$t^2+2xt-2t-2xy+2\gamma \leq 0.$$ For $x$ and $y$ fixed we can decrease $t$ decreasing $f$ by this, until $t$ become $0$ or the last constraint will become an equality. $\endgroup$ May 12, 2020 at 9:30
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    $\begingroup$ @Steve using the fact that $\lim_{n\to\infty} n(x^{\frac{1}{n}}-1)=\ln(x)$ so if we choose an arbitrary $n$ the function $f(x,y,z)$ becomes : $n[(2^{\frac{1}{n}}-1)+1.5^{\frac{1}{n}}x+2^{\frac{1}{n}}y+2.5^{\frac{1}{n}}z-(x+y+z)]$ . Maybe it helps . $\endgroup$
    – Erik Satie
    May 14, 2020 at 11:29

4 Answers 4

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In this answer we solve a particular case of the problem, when $z=0$.

Now the last constraint becomes $g(x,y)=y^2+2x-x^2-2xy\le 1-2\gamma$. Since $\frac{\partial g}{\partial y}=2y(1-x)\le 0$, for $x$ fixed we can increase $y$ increasing $f$ by this, until $y$ will be bounded by a constraint $x+y\le 1$ or $g(x,y)= 1-2\gamma $. Let’s consider these cases.

1) If $x+y=1$ then the constraint $g(x,y)\le 1-2\gamma$ becomes $x^2-x+\gamma\le 0$. In order to maximize $f(x,y,z)=\log 2+\log\frac 32+\left(\log 2-\log\frac 32\right)y$ we have to maximize $y$, that is to minimize $x$. This happens when $x=\tfrac{1-\sqrt{1-4\gamma}}2$. So it seems we have to swap $x$ and $y$ in your claim.

2) If $g(x,y)= 1-2\gamma$ then $y=x\pm\sqrt{D}$, where $$D=2x^2-2x+1-2\gamma=2\left(x-\tfrac 12\right)^2+\tfrac 34-2\gamma\ge\tfrac 14.$$ Fix $x$ and look for the constrained $y$ maximizing $f$.

Let’s check when we can take the plus sign in the formula for $y$. This is allowed iff $2x+\sqrt{D}\le 1$, that is when $x\le\tfrac 12$ and $2x^2-2x+\gamma\ge 0$, that is if $x\le x_1=\tfrac{1-\sqrt{1-2\gamma}}2$. Since $\frac{\partial D}{\partial x}>0$ when $x<\tfrac 12$, $y$ increases when $x$ increases from $x$ to $x_1$. So in this case the maximum of $f$ is attained when $x=x_1$. Then $y=1-x_1$ and this is Case 1.

If $x>x_1$ then we have $y=x-\sqrt{D}$. The constraint $x+y\le 1$ becomes $x\le\tfrac 12$ or $2x^2-2x+\gamma\le 0$, that is $x_1<x\le \tfrac{1+\sqrt{1-2\gamma}}2=x_2$. We have

$$f(x,y,z)=\log 2+x\log\frac 32+(x-\sqrt{D})\log 2=h(x).$$

Then $h’(x)=\log 2+\log\tfrac 32-\tfrac{2x-1}{\sqrt{D}}\log 2$. We claim that $h’(x)>0$. This is clear when $x\le\tfrac 12$. If $x\ge\tfrac 12$ then we have to show that $(1+c)\sqrt{D}>2x-1$, where $c=\frac{\log\tfrac 32}{\log 2}$. Let’s do this.

$(1+c) \sqrt{D}>2x-1$

$(1+c)^2(2x^2-2x+1-2\gamma)>4x^2-4x+1$

Since $(1+c)^2>2.5$, it suffices to show that

$2.5(2x^2-2x+1-2\gamma)\ge 4x^2-4x+1$

$x^2-x+1.5-5\gamma\ge 0$

$\left(x-\frac 12\right)^2+1.25-5\gamma\ge 0$, which is true because $\gamma\le 0.25$.

Thus $h$ increases when $x$ increases so a maximum of $f$ is attained when $x=x_2$. Then $y=1-x_2$ and this is Case 1 again.

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    $\begingroup$ good ${}{}{}{}{}$ $\endgroup$ May 9, 2020 at 14:42
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    $\begingroup$ @AlexRavsky Thank you for helping me out. I see that $(x,y+z, 0) \in K$, but I am having trouble seeing why $f(x,y+z, 0) \geq f(x,y,z)$. $\endgroup$
    – Steve
    May 11, 2020 at 17:11
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    $\begingroup$ @AlexRavsky If it helps, one of the ideas I had was to use the concavity of $\log(x)$ and bound it above by some secant line or some tangent line. This leads one to find a relationship between the variables that would suffice to show that $f(x,y,z)\leq \log(12)/2+\frac{\sqrt{1-4\gamma}}{2}\log(4/3)$. $\endgroup$
    – Steve
    May 12, 2020 at 3:56
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(New solution)

The maximum is $$\ln 2 + \tfrac{1-\sqrt{1-4\gamma}}{2}\ln \tfrac{3}{2} + \tfrac{1+\sqrt{1-4\gamma}}{2}\ln 2$$ at $x = \frac{1-\sqrt{1-4\gamma}}{2}$, $y = \frac{1+\sqrt{1-4\gamma}}{2}$, and $z = 0$. Let us prove it.

We first give the following auxiliary result. The proof is given at the end.

Fact 1: At optimum, either $x+y+z = 1$ or $z = 0$.

Let us proceed. Let $f = f(x, y, z) = x\ln \frac{3}{2} + y\ln 2 + z\ln \frac{5}{2}$. From Fact 1, we split into two cases:

1) $x + y + z = 1$: The constraint $(y+z)^2 + 2x - x^2 - 2xy \le 1 - 2\gamma$ becomes $xy \ge \gamma$. By using $z = 1 - x - y$, we have $$f = f(x, y) = x\ln \tfrac{3}{2} + y\ln 2 + (1-x-y)\ln \tfrac{5}{2} = \ln \tfrac{5}{2} - x\ln \tfrac{5}{3} - y\ln \tfrac{5}{4}.$$ The constraints are: $x\ge 0, y\ge 0, x + y \ge \frac{4}{5}, x+y \le 1$, and $xy \ge \gamma$.

We claim that, at optimum, $xy = \gamma$. Indeed, suppose $xy > \gamma$ at optimum (clearly, $x>\gamma$ and $\gamma < y < 1 - \gamma$), there exists $0 < \epsilon$ such that $(x - \epsilon, y + \epsilon)$ is feasible and $f(x - \epsilon, y + \epsilon) > f(x, y)$, which contradicts the optimality of $(x,y)$.

From $xy = \gamma$, we have $x + y \ge 2\sqrt{xy} = 2\sqrt{\gamma} \ge 2\sqrt{\frac{6}{25}} > \frac{4}{5}$. From $y = \frac{\gamma}{x}$ and $x + y \le 1$, we have $\frac{1-\sqrt{1-4\gamma}}{2}\le x \le \frac{1+\sqrt{1-4\gamma}}{2}$. Then, we have $$f = f(x) = \ln \tfrac{5}{2} - x\ln \tfrac{5}{3} - \tfrac{\gamma}{x}\ln \tfrac{5}{4} .$$ The constraints are: $\frac{1-\sqrt{1-4\gamma}}{2}\le x \le \frac{1+\sqrt{1-4\gamma}}{2}$. We have \begin{align} f'(x) &= - \ln \tfrac{5}{3} + \tfrac{\gamma}{x^2}\ln \tfrac{5}{4}\\ &\le - \ln \tfrac{5}{3} + \gamma(\tfrac{1-\sqrt{1-4\gamma}}{2})^{-2}\ln \tfrac{5}{4}\\ &= - \ln \tfrac{5}{3} + \gamma^{-1}(\tfrac{1+\sqrt{1-4\gamma}}{2})^2 \ln \tfrac{5}{4}\\ &\le - \ln \tfrac{5}{3} + (\tfrac{6}{25})^{-1}(\tfrac{1+\sqrt{1-4\cdot 6/25}}{2})^2 \ln \tfrac{5}{4}\\ &< 0 \end{align} for $\frac{1-\sqrt{1-4\gamma}}{2}\le x \le \frac{1+\sqrt{1-4\gamma}}{2}$. Thus, we have $$f = f(x) \le f(\tfrac{1-\sqrt{1-4\gamma}}{2}) = \ln \tfrac{5}{2} - \tfrac{1-\sqrt{1-4\gamma}}{2}\ln \tfrac{5}{3} - \tfrac{1+\sqrt{1-4\gamma}}{2}\ln \tfrac{5}{4}$$ with equality if $x = \frac{1-\sqrt{1-4\gamma}}{2}$, $y = \frac{1+\sqrt{1-4\gamma}}{2}$, and $z = 0$.

2) $z = 0$: This case was solved by other users. Although I have my solution, I will not give it since it is not better than other users' solutions.

We are done.

$\phantom{2}$

Proof of Fact 1: Assume, for the sake of contradiction, that $x + y + z < 1$ and $z > 0$, at optimum. Let $x_1 = x + 3t, y_1 = y+t, z_1 = z - 2t$ for $0 < t < \min(\frac{z}{2}, \frac{1-x-y-z}{2})$.

We have $x_1>0$, $y_1 > 0$, $z_1 > 0$, $x_1+y_1+z_1 = x+y+z + 2t < 1$ and $x_1+y_1 = x+y + 4t > \frac{4}{5}$. Also, we have \begin{align} &\big[(y+z)^2 + 2x - x^2 - 2xy\big] - \big[(y_1+z_1)^2 + 2x_1 - x_1^2 - 2x_1y_1\big]\\ =\ & 2t(7t+4x+4y+z-3)\\ >\ & 0 \end{align} where we have used $x + y \ge \frac{4}{5} > \frac{3}{4}$. Thus, $(y_1+z_1)^2 + 2x_1 - x_1^2 - 2x_1y_1 \le 1 - 2\gamma$. Thus, $(x_1, y_1, z_1)$ is feasible.

On the other hand, we have \begin{align} &\big[\ln 2 + x_1\ln \tfrac{3}{2} + y_1\ln 2 + z_1\ln \tfrac{5}{2}\big] - \big[\ln 2 + x\ln \tfrac{3}{2} + y\ln 2 + z\ln \tfrac{5}{2}\big] \\ =\ & t \ln \tfrac{27}{25}\\ >\ & 0. \end{align} However, this contradicts the optimality of $(x, y, z)$. We are done.

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  • $\begingroup$ @River_Li thank you for your response. Why does the fact that $f(x,y,z)$ strictly increases with respect to $z$ imply that equality must hold for one of those constraints? $\endgroup$
    – Steve
    May 14, 2020 at 2:47
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    $\begingroup$ @Steve Suppose two constraints are " < ", then we can increase $z$. My solution is too complicated. I may consider another solution. $\endgroup$
    – River Li
    May 14, 2020 at 2:57
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    $\begingroup$ @Steve I gave a new solution. $\endgroup$
    – River Li
    May 15, 2020 at 6:58
  • $\begingroup$ @River_Li By the way, I want to say that your proof of Fact 1 was clever. I would not have thought of that. This was a very crucial fact. Thank you. $\endgroup$
    – Steve
    May 19, 2020 at 6:57
  • $\begingroup$ @Steve You are welcome. $\endgroup$
    – River Li
    May 19, 2020 at 7:30
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I cannot prove that the maximum value is $$M:=\frac{\log(12)}{2}+\frac{\log(4/3)}{2}\sqrt{1-4\gamma}$$ but let me show one possible way to prove that the maximum value is $M$.


If $x=0$, then $\frac 45\le y\le y+z\leq \sqrt{1-2\gamma}\implies \frac 45\le \sqrt{1-2\gamma}\implies \gamma\le 0.18$ which is impossible. So, we have $x\gt 0$.

In order for $z$ satisying $$z^2+2yz+y^2+2x-x^2-2xy+2\gamma-1\leq 0\tag1$$ (seen as a quadratic inequality on $z$) to exist, it is necessary that the discriminant $\ge 0$, i.e. $$y\ge\frac{2x-x^2+2\gamma-1}{2x}\tag2$$

Under $(2)$, we see that $(1)$ is equivalent to

$$-y-\sqrt{-2x+x^2+2xy-2\gamma+1}\le z\le -y+\sqrt{-2x+x^2+2xy-2\gamma+1}\tag3$$

In order for $z$ satisfying $(3)$ and $0\le z\le 1-x-y$ to exist, it is necessary that $-y+\sqrt{-2x+x^2+2xy-2\gamma+1}\ge 0$, i.e. $$x-\sqrt{2x^2-2x-2\gamma+1}\le y\le x+\sqrt{2x^2-2x-2\gamma+1}\tag4$$

It follows from $(3)$ and $0\le z\le 1-x-y$ that $$0\le z\le\min\bigg(1-x-y,-y+\sqrt{-2x+x^2+2xy-2\gamma+1}\bigg)$$ where $$\min\bigg(1-x-y,-y+\sqrt{-2x+x^2+2xy-2\gamma+1}\bigg)$$ $$=\begin{cases}1-x-y&\text{if $\quad xy\ge \gamma$}\\\\-y+\sqrt{-2x+x^2+2xy-2\gamma+1}&\text{if $\quad xy\le \gamma$}\end{cases}$$


Case 1 : $xy\ge\gamma$

We get $0\le z\le 1-x-y$ from which we have $$\begin{align}f(x,y,z)&\le f(x,y,1-x-y)=\log(5)-\log(5/3)x-\log(5/4)y:=g(x,y)\end{align}$$ So, we want to maximize $g(x,y)$ under the condition that $$x\gt 0, y\geq 0, 0\le 1-x-y,x+y\geq \frac 45,(2),(4),xy\ge\gamma$$

i.e. $$\frac{1-\sqrt{1-4\gamma}}{2}\le x\le\frac{1+\sqrt{1-4\gamma}}{2},\frac{\gamma}{x}\le y\leq 1-x$$

So, we get $$g(x,y)\le g\bigg(x,\frac{\gamma}{x}\bigg)=\log(5)-\log(5/3)x-\log(5/4)\frac{\gamma}{x}:=h(x)$$ where $h'(x)=\frac{-\log(5/3)x^2+\log(5/4)\gamma}{x^2}$ with $h'(x)=0\iff x=\sqrt{\frac{\log(5/4)\gamma}{\log(5/3)}}$. Since $\sqrt{\frac{\log(5/4)\gamma}{\log(5/3)}}\lt \frac{1-\sqrt{1-4\gamma}}{2}$, we see that $h'(x)\lt 0$ and that $h(x)$ is decreasing, so we get $$f(x,y,z)\le h(x)\le h\bigg(\frac{1-\sqrt{1-4\gamma}}{2}\bigg)=M$$ which is attained when $(x,y,z)=\bigg(\frac{1-\sqrt{1-4\gamma}}{2},\frac{1+\sqrt{1-4\gamma}}{2},0\bigg)$.


Case 2 : $xy\le\gamma$

We get $0\le z\le -y+\sqrt{-2x+x^2+2xy-2\gamma+1}$ from which we have $$\small\begin{align}f(x,y,z)&\le f\bigg(x,y,-y+\sqrt{-2x+x^2+2xy-2\gamma+1}\bigg) \\\\&=\log(2)+\log(3/2)x-\log(5/4)y+\log(5/2)\sqrt{-2x+x^2+2xy-2\gamma+1}:=i(x,y)\end{align}$$

So, we want to maximize $i(x,y)$ under the condition that $$x\gt 0, y\geq 0, 0\le 1-x-y,x+y\geq \frac 45,(2),(4),xy\le\gamma$$

i.e. $$ \frac{3-\sqrt{27-100\gamma}}{10}\le x\le \frac{1+\sqrt{1-4\gamma}}{2},$$ $$\max\bigg(\frac 45-x,x-\sqrt{2x^2-2x-2\gamma+1}\bigg)\le y\le \min\bigg(\frac{\gamma}{x},x+\sqrt{2x^2-2x-2\gamma+1}\bigg)$$

Now, $\frac{\partial i}{\partial y}=-\log(5/4)+\frac{\log(5/2)x}{\sqrt{-2x+x^2+2xy-2\gamma+1}}$ which is decreasing with $\frac{\partial i}{\partial y}=0\iff y=\frac{(1-c)x^2+2cx+2c\gamma-c}{2cx}$ where $c=\bigg(\frac{\log(5/4)}{\log(5/2)}\bigg)^2$.

Here, we can separate it into the following six cases :

Case 2-1 : $0.24\le r\le 0.2475$

Case 2-1-1 : $\frac{3-\sqrt{27-100\gamma}}{10}\le x\le \frac{1-\sqrt{1-4\gamma}}{2}$

$$-x+\frac 45\le y\le x+\sqrt{2x^2-2x-2\gamma+1}$$

Case 2-1-2 : $\frac{1-\sqrt{1-4\gamma}}{2}\le x\le\frac{3+\sqrt{27-100\gamma}}{10}$

$$-x+\frac 45\le y\le\frac{\gamma}{x}$$

Case 2-1-3 : $\frac{3+\sqrt{27-100\gamma}}{10}\le x\le{\frac{1+\sqrt{1-4\gamma}}{2}}$

$$x-\sqrt{2x^2-2x-2\gamma+1}\le y\le\frac{\gamma}{x}$$

Case 2-2 : $0.2475\le r\le 0.25$

Case 2-2-1 : ${\frac{3-\sqrt{27-100\gamma}}{10}}\le x\le \frac{3+\sqrt{27-100\gamma}}{10}$

$$-x+\frac 45\le y\le x+\sqrt{2x^2-2x-2\gamma+1}$$

Case 2-2-2 : $\frac{3+\sqrt{27-100\gamma}}{10}\le x\le\frac{1-\sqrt{1-4\gamma}}{2}$

$$x-\sqrt{2x^2-2x-2\gamma+1}\le y\le x+\sqrt{2x^2-2x-2\gamma+1}$$

Case 2-2-3 : $\frac{1-\sqrt{1-4\gamma}}{2}\le x\le{\frac{1+\sqrt{1-4\gamma}}{2}}$

$$x-\sqrt{2x^2-2x-2\gamma+1}\le y\le \frac{\gamma}{x}$$


Now, let $$\begin{align}F_1(x)&:=i\bigg(x,\frac 45-x\bigg)\\\\ F_2(x)&:=i\bigg(x,x-\sqrt{2x^2-2x-2\gamma+1}\bigg)\\\\ F_3(x)&:=i\bigg(x,\frac{\gamma}{x}\bigg)\\\\ F_4(x)&:=i\bigg(x,x+\sqrt{2x^2-2x-2\gamma+1}\bigg)\\\\ F_5(x)&:=i\bigg(x,\frac{(1-c)x^2+2cx+2c\gamma-c}{2cx}\bigg)\end{align}$$

Also, let $x_i$ be such that $F_i'(x_i)=0$ for each $i=1,2,\cdots, 5$.

Now, if it is true that, for $0.24\le \gamma\le 0.25$, $$F_1\bigg(\frac{3-\sqrt{27-100\gamma}}{10}\bigg)\le M,F_1\bigg(\frac{3+\sqrt{27-100\gamma}}{10}\bigg)\le M,F_1(x_1)\le M,$$

$$F_2\bigg(\frac{3+\sqrt{27-100\gamma}}{10}\bigg)\le M,F_2\bigg({\frac{1+\sqrt{1-4\gamma}}{2}}\bigg)\le M,F_2(x_2)\le M,$$

$$F_3\bigg(\frac{1-\sqrt{1-4\gamma}}{2}\bigg)\le M,F_3\bigg({\frac{1+\sqrt{1-4\gamma}}{2}}\bigg)\le M,F_3(x_3)\le M,$$

$$F_4\bigg(\frac{1-\sqrt{1-4\gamma}}{2}\bigg)\le M,F_4\bigg({\frac{1+\sqrt{1-4\gamma}}{2}}\bigg)\le M,F_4(x_4)\le M,$$

$$F_5\bigg({\frac{3-\sqrt{27-100\gamma}}{10}}\bigg)\le M,F_5\bigg(\frac{1+\sqrt{1-4\gamma}}{2}\bigg)\le M,F_5(x_5)\le M$$

then we can say that the maximum value is $M$. (I have to say that the calculations are very tedious.)

(That at least one of them is larger than $M$ does not necessarily mean that $M$ is not the maximum value. If at least one of them is larger than $M$, then we have to consider the case(s) more carefully.)

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Note that $$f(x,y,z)=\log2+(x+y+z)\log\frac52-(x+y)\log\frac53+y\log\frac43$$ which is maximised when $z=0$ since the negative term $-(x+y)\ge z-1$ will be minimised and at the same time $y\le1-x-z$ will be maximised. Then it is just a case of maximising $$f(x,y,0)=\log2+x\log\frac32+y\log2$$ subject to $4/5\le x+y\le1$ and $xy\ge\gamma$, or just $\gamma/y\le x\le1-y$. Solving the quadratic yields $$a:=\frac{1-\sqrt{1-4\gamma}}2\le x,y\le\frac{1+\sqrt{1-4\gamma}}2:=b$$ and $f$ will be maximised at the endpoints of $x,y$. Recalling that $x+y\le1$, the only possibilities are $(x,y)=(a,b)$ and $(b,a)$. As $\log2>\log3/2$, the maximum occurs when $y$ takes the positive root, so \begin{align}\max f(x,y,z)=f(a,b,0)&=\log2+\frac12\log\frac32-\frac{\sqrt{1-4\gamma}}2\log\frac32+\frac12\log2+\frac{\sqrt{1-4\gamma}}2\log2\\&=\frac12\log12+\frac{\sqrt{1-4\gamma}}2\log\frac43\end{align} whose overall maximum is $0.1\log331776$ at $\gamma=0.24$.

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    $\begingroup$ I don’t see why your arguments show that $f(x,y,z)$ can be maximized when $z=0$. Variables $x$, $y$, and $z$ are related via the constraints. Given $(x,y,z)$ satisfying the constraints how can we find $(x’,y’,0)$, also satisfying the constraints, such that $f(x’,y’,0)\ge f(x,y,z)$? $\endgroup$ May 12, 2020 at 18:17
  • $\begingroup$ @AlexRavsky We want to maximise $x+y+z$, minimise $-(x+y)$ and maximise $y$ all at the same time to yield a maximum $f$. This can only be done when $z=0$ because of the constraint $x+y+z\le1$. Any other $z>0$ the value of $-(x+y)$ will not be minimised, and the value of $y\le1-x-z$ will not be maximised. $\endgroup$
    – TheSimpliFire
    May 12, 2020 at 19:20
  • $\begingroup$ @TheSimpliFire Do you mean to maximize $-(x+y)$ rather than minimize $-(x+y)$? $\endgroup$
    – River Li
    May 13, 2020 at 0:35
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    $\begingroup$ Values of $x+y+z$, $-(x+y)$, and $y$ are not independent, they are related via the constraints. Why we can maximize the first and the third and minimize the second at the same time? Maybe it happens that for some$ z>0$ the value of $−(x+y)$ is not minimized or the value of $y$ is not maximized, but due to the constraints (other than $x+y+z\le 1$) we cannot increase $f$ by decreasing $z$ and increasing $y$ or $x$. $\endgroup$ May 14, 2020 at 3:32

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