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

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