<s>[INST] Maximize $\log(2)+\log(3/2)x+\log(2)y+\log(5/2)z$ if $x+y+z\leq 1$ and $(y+z)^2+2x-x^2-2xy\leq 1-2\gamma$, $0.24 \leq \gamma \leq 0.25$ 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.
 [/INST] 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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