Sí, es cierto que 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}
donde c_0\approx 0.358678 es una raíz de c^4-8 c^3+22 c^2-32 c+9.
Vamos
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}
A continuación,
\frac{\partial f(a,b,c)}{\partial a}=\frac{g(a)}{(3-c-3a+ac+b)^2}
donde
\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}
Desde
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
tenemos que
a_0=\frac{-(-b+c-3)-\sqrt{P}}{3-c}\tag1
donde
P=b^2 c-2 b^2-2 b c+6 b+c^3-3 c^2-2 c+6
Ahora,
\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}
Multiplicar por \frac{(3-c)^2}{(3-c)^2} da
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)}
Usando ese a_0^2(3-c)=-2(-b+c-3)a_0-b^2-c^2+c-1 da
\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}
Así,
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
Aquí, vamos a
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
A continuación,
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}}
Desde 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}
Aquí, tenga en cuenta que
0\lt b_-\lt 1\lt b_+
donde
b_{\pm}=\frac{3-c\pm\sqrt{(c-3)^2-(2-c)(c^2-c+1)}}{2-c}
dando
f(a,b,c)\le f(a_0,b,c)\le f(a_0,b_-,c)
Ahora, utilizando
(2-c){b_-}^2+2(c-3){b_-}=-c^2+c-1
da
\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}
Así
\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}
Aquí, vamos a
h(c):=\frac{\sqrt{c^3-2 c^2-3 c+7}-1}{2-c}
A continuación,
\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}
Desde -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}
Por lo tanto, c_0\approx 0.358678 es la única raíz real en [0,1] de
c^4-8 c^3+22 c^2-32 c+9
y
\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}
de la siguiente manera.