Upper bound of product

Question

If $x \in [a,b]$, then I want to find the optimal upper bound of the product $$ |x - a| |x - b| \leq M $$ It seems obvious that $$ |x - a| |x - b| \leq |b-a|^2 $$ however it seems that the optimal upper bound is in fact $$ |x-a||x-b| \leq \frac{|b-a|^2}{4} $$ Does anyone know how to prove this?

Answer 1

First using the arithmetic mean - geometric mean inequality, we get that

$$ |x-a||x-b| \le \left(\frac{|x-a|+|x-b|}{2}\right)^2.$$

Now since $x \in [a,b]$ we have that $|x-a|+|x-b| = |b-a|$. Thus, we have that

$$ |x-a||a-b| \le \frac{|b-a|^2}{4}.$$

---

This bound is tight, as can be seen by picking $x = (b+a)/2$. Then

$$|x-a||x-b| = \frac{|b-a|}{2}\frac{|a-b|}{2} = \frac{|b-a|^2}{4}.$$

Answer 2

In order to prove a statement $A$ it can be worthwhile to search for a stronger statement $B$ that is simpler to establish and implies statement $A$.

The task at hand requires knowledge about the range of the function $$f : [a,b] \rightarrow \mathbb{R}$$ given by $$ f(x) = (x-a)(x-b).$$ We have $$f(a) = f(b) = 0.$$ It is clear that $f$ is differentiable and $$f'(x) = 2x - (a+b).$$ We have $$f'(x) = 0$$ if and only if $$x= \frac{a+b}{2} =: \mu$$ Since $f$ is continuous and $$f(\mu) = -\frac{(b-a)^2}{4} =: -M$$ the range of $f$ is the interval $[-M,0]$ and the desired result is immediate.

---

Obviously, simplicity is a matter of perspective and depends on the intended audience, which is why it is so valuable to have different proofs for the same result.

Answer 3

Since $a \leq x \leq b$, we have,

$|x-a||x-b|=(x-a)(b-x)$

$=\left(\frac{x-a+b-x}{2}\right)^2-\left(\frac{x-a-b+x}{2}\right)^2$, by the identity $pq=\left(\frac{p+q}{2}\right)^2-\left(\frac{p-q}{2}\right)^2$

$=\left(\frac{b-a}{2}\right)^2-\left(\frac{2x-a-b}{2}\right)^2 \leq \frac{(b-a)^2}{4}$,

the equality holds when $2x=a+b$ i.e., $x=\frac{a+b}{2}$.