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

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.

• It would be easier if you provided the rest of the proof. Without it, it becomes harder to understand where the author is going and hence harder to explain you why that should make sense – RGS Jan 20 '17 at 15:45
• @RSerrao Will do. – The Pointer Jan 20 '17 at 15:45
• Thanks; you may want to check any of the answers below. You are provided with various proofs; I took the liberty of showing where that inequality comes from; – RGS Jan 20 '17 at 15:57

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?

• Thanks for the response. How did you get $-b \le ax + \dfrac{-b}{2}$ from $\dfrac{-b}{2} \le ax$? – The Pointer Jan 20 '17 at 16:03
• @ThePointer I subtracted $\frac{-b}{2}$ from both sides – RGS Jan 20 '17 at 16:05
• That makes sense. Thank you very much for your assistance. – The Pointer Jan 20 '17 at 16:05
• @ThePointer you are welcome. I am going to tell you how to cheat a little: when presented with things like this, try to go the other way around. the author had some assumptions and got to a result he wanted to use (in this case, the inequality); you can always start with the result we want, and try to derive the assumptions. If along the way you only do equivalent steps, then you will be able to go all the way back again, making the results appear from the assumptions. Did this make sense? – RGS Jan 20 '17 at 16:09
• Yes, I think this is good advice. I found this derivation particularly troublesome, since the inequality seemed to be far detached from earlier assumptions. Indeed, I think your method would have proved more effective. – The Pointer Jan 20 '17 at 16:12

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.

• that is indeed a proof, but it does not address one of the issues of the OP which is to understand the inequality provided – RGS Jan 20 '17 at 15:48

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

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

• likewise, this does not seem to address the main problem of the OP – RGS Jan 20 '17 at 15:57