TITLE: Maximizing $\frac{(1-a)(2+b+ab-b^2-c^2)}{(1-a)(3-c)+b}$
QUESTION [5 upvotes]: Let $0<\varepsilon<0.01$, and consider the values $a_0,b_0,c_0$ with $a_0\in[0,1-\varepsilon]$ and $b_0,c_0\in[0,1]$ that maximize
$$\frac{(1-a)(2+b+ab-b^2-c^2)}{(1-a)(3-c)+b}.$$
Is it true that $a_0=b_0$?
I've verified this by software for $\varepsilon=0.001$. The function is maximized at $a_0=b_0\approx 0.153$ and $c_0\approx 0.359$. If we allow $a=1$, then along with $b=0$ the function is $\frac00$. And obviously the function is not symmetric in $a,b$, so we cannot apply a symmetry argument.

REPLY [6 votes]: Yes, it is true that $a_0=b_0$ : 
$$a_0=b_0=\frac{3-c_0-\sqrt{c_0^3-2c_0^2-3c_0+7}}{2-c_0}$$
where $c_0\approx 0.358678$ is a root of $c^4-8 c^3+22 c^2-32 c+9$.

Let 
$$f(a,b,c):=\frac{(1-a)(2+b+ab-b^2-c^2)}{(1-a)(3-c)+b}=\frac{(1-a)(2+b+ab-b^2-c^2)}{3-c-3a+ac+b}$$
Then,
$$\frac{\partial f(a,b,c)}{\partial a}=\frac{g(a)}{(3-c-3a+ac+b)^2}$$
where
$$\begin{align}g(a)&=(-(2+b+ab-b^2-c^2)+(1-a)b)(3-c-3a+ac+b)\\&\qquad -(1-a)(2+b+ab-b^2-c^2)(-3+c)\\&=b((3-c)a^2+2(-b+c-3)a+b^2+c^2-c+1)\end{align}$$
Since
$$3-c\gt 0,\quad g(0)=b(b^2+c^2-c+1)\gt 0,\quad g(1)=b(b(b-2)+c^2-2)\lt 0$$
we have that 
$$a_0=\frac{-(-b+c-3)-\sqrt{P}}{3-c}\tag1$$
where
$$P=b^2 c-2 b^2-2 b c+6 b+c^3-3 c^2-2 c+6$$
Now,
$$\begin{align}f(a,b,c)\le f(a_0,b,c)&=\frac{(1-a_0)(2+b+a_0b-b^2-c^2)}{3-c-3a_0+a_0c+b}\\&=\frac{-ba_0^2+(-2+b^2+c^2)a_0+2+b-b^2-c^2}{3-c-a_0(3-c)+b}\end{align}$$
Multiplying it by $\frac{(3-c)^2}{(3-c)^2}$ gives
$$f(a_0,b,c)=\frac{(3-c)(-ba_0^2(3-c)+(-2+b^2+c^2)a_0(3-c)+(2+b-b^2-c^2)(3-c))}{(3-c)^2(3-c-a_0(3-c)+b)}$$
Using that $a_0^2(3-c)=-2(-b+c-3)a_0-b^2-c^2+c-1$ gives
$$\begin{align}&(3-c)(-ba_0^2(3-c)+(-2+b^2+c^2)a_0(3-c)+(2+b-b^2-c^2)(3-c))\\&=(3-c)(-b(-2(-b+c-3)a_0-b^2-c^2+c-1)+(-2+b^2+c^2)a_0(3-c))\\&\qquad +(2+b-b^2-c^2)(3-c)^2\\&=(3-c)a_0(-b^2 c+b^2+2 b c-6 b-c^3+3 c^2+2 c-6)\\&\qquad +(3-c)(b^3+b^2 c-3 b^2+b c^2-2 b c+4 b+c^3-3 c^2-2 c+6)\\&=(b-c+3-\sqrt{P})(-b^2 c+b^2+2 b c-6 b-c^3+3 c^2+2 c-6)\\&\qquad +(3-c)(b^3+b^2 c-3 b^2+b c^2-2 b c+4 b+c^3-3 c^2-2 c+6)\\&=(P+b^2)\sqrt{P}-2bP\end{align}$$
So,
$$f(a_0,b,c)=\frac{(P+b^2)\sqrt P-2bP}{(3-c)^2\sqrt{P}}=\frac{P+b^2-2b\sqrt P}{(3-c)^2}=\left(\frac{\sqrt P-b}{3-c}\right)^2$$
Here, let 
$$g(b):=\sqrt P-b=\sqrt{b^2 c-2 b^2-2 b c+6 b+c^3-3 c^2-2 c+6}-b$$
Then,
$$g'(b)=\frac{bc-2b-c+3-\sqrt{b^2 c-2 b^2-2 b c+6 b+c^3-3 c^2-2 c+6}}{\sqrt{b^2 c-2 b^2-2 b c+6 b+c^3-3 c^2-2 c+6}}$$
Since $bc-2b-c+3\gt 0$, 
$$\begin{align}g'(b)\ge 0&\iff bc-2b-c+3\ge \sqrt{b^2 c-2 b^2-2 b c+6 b+c^3-3 c^2-2 c+6}\\&\iff (bc-2b-c+3)^2\ge b^2 c-2 b^2-2 b c+6 b+c^3-3 c^2-2 c+6\\&\iff (c-2)(c-3)b^2-2(c-3)^2b-(c-3)(c^2-c+1)\ge 0\\&\iff (2-c)b^2+2(c-3)b+c^2-c+1\ge 0\end{align}$$
Here, note that
$$0\lt b_-\lt 1\lt b_+$$
where
$$b_{\pm}=\frac{3-c\pm\sqrt{(c-3)^2-(2-c)(c^2-c+1)}}{2-c}$$
giving
$$f(a,b,c)\le f(a_0,b,c)\le f(a_0,b_-,c)$$
Now using
$$(2-c){b_-}^2+2(c-3){b_-}=-c^2+c-1$$
gives
$$\begin{align}P_{b=b_-}&=-(2-c){b_-}^2-2(c-3){b_-}+c^3-3 c^2-2 c+6\\&=c^2-c+1+c^3-3 c^2-2 c+6\\&=c^3-2c^2-3c+7\end{align}$$
So
$$\begin{align}f(a_0,b_-,c)&=\left(\frac{\sqrt{c^3-2c^2-3c+7}-\frac{3-c-\sqrt{c^3-2 c^2-3 c+7}}{2-c}}{3-c}\right)^2\\&=\left(\frac{(2-c)\sqrt{c^3-2c^2-3c+7}-(3-c-\sqrt{c^3-2 c^2-3 c+7})}{(2-c)(3-c)}\right)^2\\&=\left(\frac{(3-c)\left(\sqrt{c^3-2 c^2-3 c+7}-1\right)}{(2-c)(3-c)}\right)^2\\&=\left(\frac{\sqrt{c^3-2 c^2-3 c+7}-1}{2-c}\right)^2\end{align}$$
Here, let
$$h(c):=\frac{\sqrt{c^3-2 c^2-3 c+7}-1}{2-c}$$
Then,
$$\begin{align}h'(c)&=\frac{\frac{3c^2-4c-3}{2\sqrt{c^3-2c^2-3c+7}}(2-c)+\sqrt{c^3-2 c^2-3 c+7}-1}{(2-c)^2}\\&=\frac{-c^3+6 c^2-11 c+8-2\sqrt{c^3-2c^2-3c+7}}{2(2-c)^2\sqrt{c^3-2 c^2-3 c+7}}\end{align}$$
Since $-c^3+6 c^2-11 c+8\gt 0$,
$$\begin{align}h'(c)\ge 0&\iff -c^3+6 c^2-11 c+8\ge 2\sqrt{c^3-2c^2-3c+7}\\&\iff (-c^3+6c^2-11c+8)^2\ge 4(c^3-2c^2-3c+7)\\&\iff (c-2)^2 (c^4-8 c^3+22 c^2-32 c+9)\ge 0\\&\iff c^4-8 c^3+22 c^2-32 c+9\ge 0\end{align}$$
Therefore, $c_0\approx 0.358678$ is the only real root in $[0,1]$ of
$$c^4-8 c^3+22 c^2-32 c+9$$
and
$$\begin{align}a_0&=\frac{b_0-c_0+3-\sqrt{c_0^3-2c_0^2-3c_0+7}}{3-c_0}\\&=\frac{\frac{3-c_0-\sqrt{c_0^3-2c_0^2-3c_0+7}}{2-c_0}-c_0+3-\sqrt{c_0^3-2c_0^2-3c_0+7}}{3-c_0}\\&=\frac{3-c_0-\sqrt{c_0^3-2c_0^2-3c_0+7}}{2-c_0}\\&=b_-\\&=b_0\end{align}$$
follows.