Upper bound for absolute value integral

Question

Q: For distinct $x_0, x_1, \cdots, x_n \in [0,1]$, show that

\begin{align*} \int_0^1 |(x-x_0)(x-x_1)\cdots(x-x_n)| \mathrm{d} x \leq 1 \end{align*}

What I have so far: we can assume $x_0 < x_1 < \cdots < x_n$. Then we can partition the interval and the LHS can be written as \begin{align*} \int_0^{x_0} &(-1)^{n+1}(x-x_0)(x-x_1)\cdots(x-x_n) \mathrm{d} x +\int_{x_0}^{x_1} (-1)^{n}(x-x_0)(x-x_1)\cdots(x-x_n) \mathrm{d} x \\ &+ \cdots + \int_{x_{n-1}}^{x_n} (-1)(x-x_0)(x-x_1)\cdots(x-x_n) \mathrm{d} x + \int_{x_n}^{1}(x-x_0)(x-x_1)\cdots(x-x_n) \mathrm{d} x \end{align*}

From here, I'm stuck; I think I'm overthinking things? Any help would be appreciated!

Answer 1

Rather, $|x-x_{i}|\leq 1-0=1$ for all $i=0,1,...,n$ and $x\in[0,1]$, so \begin{align*} \int_{0}^{1}|x-x_{0}|\cdots|x-x_{n}|dx\leq\int_{0}^{1}1^{n+1}dx=1. \end{align*}

Note that for $x\in[0,1]$, either $x\geq x_{i}$ or $x<x_{i}$, in the former case, $|x-x_{i}|=x-x_{i}\leq 1-0=1$, in the later case, $|x-x_{i}|=x_{i}-x\leq 1-0=1$ still.

Answer 2

Take the max of each term in the product. It will be $|x_i|$ or $|1-x_i|$, depending on whether $x_i$ is larger than $\frac{1}{2}$. Either way, denoting

$$M_i = \max\left(|x_i|,|1-x_i|\right)$$

each $M_i$ is less than $1$. This leaves us with

$$\int_0^1 \prod_{i=0}^n |x-x_i| dx \leq (1-0)\prod_{i=0}^n M_i \leq 1$$

by triangle inequality on the integral.