If $a$, $b$, and $c$ are real numbers for which $a < 0$, then $x^* = \dfrac{−b}{2a}$ is a maximizer of $f(x) = ax^2 + bx + c$

Question

If $a$, $b$, and $c$ are real numbers for which $a < 0$, then $x^* = \dfrac{−b}{2a}$ is a maximizer of $f(x) = ax^2 + bx + c$.

The author gives the following proof:

Let $x$ be a real number. If $x^* \ge x$, then $x^* − x \ge 0$ and $a(x^* + x) + b \ge 0$. So, $(x^* − x)[a(x^* + x)+b] \ge 0$. On multiplying the term $x^* − x$ through, rearranging terms, and adding $c$ to both sides, one obtains that $a(x^*)^2 + bx^* + c \ge ax^2 + bx + c$. A similar argument applies when $x^* < x$.

I am confused as to where the inequality $a(x^* + x) + b \ge 0$ comes from. I have analysed the proposition, but I cannot see how such an inequality can be derived.

I would greatly appreciate it if the knowledgeable members of MSE could please take the time to clarify this.

Thank you.

Answer 1

Note that $x^* = \frac{-b}{2a}$ and that we assumed $x^* \geq x$:

$$x^* \geq x \iff \frac{-b}{2a} \geq x \iff \frac{-b}{2} \leq ax \iff -b \leq ax + \frac{-b}{2} \iff\\ a\left(x + \frac{-b}{2a}\right) \geq -b \iff a(x + x^*) + b \geq 0$$

I would argue that the way one gets to this inequality is not very straightforwad. It is easier to show the inequality the other way around: starting with $a(x + x^+) + b \geq 0$ and reaching to the point where $x^* \geq x$.

Can you understand the rest of the argument?

Answer 2

we got $$f'(x)=2ax+b$$ and the solution of this equation is $$x=-\frac{b}{2a}$$ since $$f''(x)=2a<0$$ we have a Maximum at this Point.

Answer 3

$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}\leq-\frac{b^2-4ac}{4a}$.

The equality occurs for $x=-\frac{b}{2a}$.