Given the nodes $x_0<…<x_n$ that are symmetric, $x_{n-j}=x_j$, how do I show that for $xin text{conv}{x_j}$

$$

|w_n(x)|=|(x-x_0)(x-x_1)…(x-x_n)|leqfrac{|x_n-x_0|^{n+1}}{2^{n+1}}

$$

I get that I can use the symmetry $x_{n-j}=x_j Leftrightarrow x-x_{n-j}=x-x_j$. Then the book does the step where I do not understand the last equality only:

$$

|(x-x_j)(x-x_{n-j})|=|(x-x_j)(x+x_{j})|=|x^2-x_j^2|leq(?)frac{1}{4}|x_n-x_0|^2

$$

How do I get this last inquality?