TITLE: Maximize $\log(2)+\log(3/2)x+\log(2)y+\log(5/2)z$ if $x+y+z\leq 1$ and $(y+z)^2+2x-x^2-2xy\leq 1-2\gamma$, $0.24 \leq \gamma \leq 0.25$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.)