If $a$ is a negative number, then $\bar{x} = \dfrac{-b}{2a}$ is a maximiser of the function $ax^2 + bx + c$: Proof Clarification.

Question

I have the following proposition:

If $a$ is a negative number, then $\bar{x} = \dfrac{-b}{2a}$ is a maximiser of the function $ax^2 + bx + c$.

The author works forwards from the hypothesis (A) and backwards from the conclusion (B):

B1: For every real number $x$, $a\bar{x}^2 + b\bar{x} + c \ge ax^2 + bx + c$

A1: A real number $x$

B2: $a\bar{x}^2 + b\bar{x} + c \ge ax^2 + bx + c$

Subtracting $ax^2 + bx + c$ from both sides of B2 and factoring out $\bar{x} - x$, it must be shown that

B3: $(\bar{x} - x)[a(\bar{x} + x) + b] \ge 0$

If $\bar{x} - x = 0$, then B3 is true. Thus you can assume that

A2: $\bar{x} - x \not= 0$

It is important to note here that you can rewrite A2 as follows so as to contain the keywords "either/or" explicitly:

A3: Either $\bar{x} - x > 0$ or $\bar{x} - x < 0$

Case 1: Assume that

A4: $\bar{x} - x > 0$

In this case you can divide both sides of B3 by the positive number $\bar{x} - x$; thus it must be shown that

B4: $a(\bar{x} + x) + b \ge 0$

Now here is where I am stuck:

Working forward from the fact that $\bar{x} = \dfrac{-b}{2a}$ and $a < 0$ (see the hypothesis), it follows from A4 that

A5: $2a\bar{x} + b > 0$

I don't see how the author got $2a\bar{x} + b > 0$? When I work forward from $\bar{x} = \dfrac{-b}{2a}$, I get

$\bar{x} = \dfrac{-b}{2a}$

$\implies \bar{x} + \dfrac{b}{2a} = 0$.

$\implies 2a\bar{x} + b = 0$, where $a < 0$.

I've gone through the proof multiple times, but I cannot see how the author possibly got $2a\bar{x} + b > 0$?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

Note that we should work from A4 to B4. Now, $$A4: \bar{x}-x >0 \implies \bar{x}>x$$

and we should use $$a(\bar{x}+x)+b \geq 0\tag{1}$$ Note that: $$\bar x > x \implies \bar x + x < 2\bar x \implies a(\bar x + x) + b < 2a\bar x+ b\tag{2}$$

What can you conclude from $(1)$ and $(2)$?