A quadratic $ax^2+bx+c$ has its roots in the interval $[0,1]$, find the maximum value of $\frac{(a-b)(2a-b)}{a(a-b+c)}$

Question

A friend of mine sent me this problem asking for help, however, I myself, need help, so here am I. I was able to find the minimum of the denominator, which was equal to $b^2/a$ if my steps were correct, using some inequalities which I got from the discriminant of the quadratic and the fact that its roots are in the interval $[0,1]$. However, I couldn't do much for the nuemrator.

Answer 1

Since $ax^2 + bx + c = 0$ has the same roots as $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$ for $a \ne 0$, this suggests writing instead $$\frac{(a-b)(2a-b)}{a(a-b+c)} = \frac{(1 - b/a)(2 - b/a)}{1 - b/a + c/a}$$ and letting $p = b/a$, $q = c/a$, for which we now seek to maximize $$f(p,q) = \frac{(1-p)(2-p)}{1-p+q}.$$ But if $r_1, r_2 \in [0,1]$ are the roots, we also have $$(x-r_1)(x-r_2) = x^2 + px + q,$$ hence $$p = -(r_1 + r_2), \quad q = r_1 r_2,$$ after which we find $$f(-r_1-r_2, r_1 r_2) = 2 + \frac{r_1}{1+r_2} + \frac{r_2}{1 + r_1}.$$ Now can you maximize this expression?