Preselection tests for the full-time training

Let $u,v$ be the roots of the quadratic equation $a{x}^2+bx+c=0$ such that $0\le u\le v\le 1$. We have the relations $b=-a(u+v)$ and $c=uv$. Therefore $$\begin{array}{rl}\frac{(a-b)(2a-b)}{a(a-b+c)}& =\frac{(1+u+v)(2+u+v)}{1+u+v+uv}\\ & =2+\frac{u}{1+v}+\frac{v}{1+u}\\ & \le 2+\frac{u}{1+u}+\frac{1}{1+u}=3\end{array}$$ Clearly, when $u=v=1$, the equality holds. Thus, $3$ is the maximum.