If we take nodes $x_j = \cos\frac{2j - 1}{2n} \pi$ we can observe that:
$$\|(x-x_1)(x-x_2)...(x-x_n)\| = \|\frac{1}{2^{n - 1}}T_n(x)\| \le \frac{1}{2^{n - 1}}$$
where we used fact, that $T_n$ is a Chebyshev polynomial, so its leading coefficient equals to $\frac{1}{2^{n - 1}}$, and for $x \in [-1, 1]$ we have that $\|T_n(x)\| \le 1$
However, this theorem applies only to nodes on interval $[-1, 1]$, and I would like to generalize this theorem for any interval $[a, b]$. I found some information how to do it, however, I don't know exactly how the proof should look like:
My work so far
Instead of taking $x_j = \cos \frac{2 j - 1}{2n}\pi$ we will take $x_j = \frac 1 2 (a + b) + \frac 1 2 (b - a)\cos(\frac{2j - 1}{2n}\pi)$. We need to observe, that in such a case $b \ge x_j \ge a$, and:
$$T_n(\frac{2x_j - (a + b)}{b - a}) = 0$$
So since numbers $x_j$ are roots of polynomial $T_n(\frac{2x_j - (a + b)}{b - a}) = 0$, we can write that:
$$\|(x - x_1)...(x-x_n)\| \le \|\frac{1}{2^{n - 1}}T_n(\frac{2x_j - (a + b)}{b - a})\| \le \frac{1}{2^{n - 1}}$$
Since $\|T_n(\frac{2x_j - (a + b)}{b - a}) = 0\| \le 1.$
However, its not exactly the desire result. What I wanted to obtain is inequality:
$$\|(x-x_1)...(x-x_n)\| \le \frac{1}{2^{n - 1}}(\frac{b - a}{2})^n$$
Could you please tell me where I'm doing the mistake?
Stay with $x_j = \cos\frac{2j - 1}{2n} \pi$ and set $z_j=c+dx_j$ to not re-use the same letter again. Then $$ (z-z_1)(z-z_2)...(z-z_n)=\prod_{j=1}^n(z-c-dx_j)=d^n\prod_{j=1}^n\left(\frac{z-c}{d}-x_j\right) \\=\frac{d^n}{2^{n-1}}\,T_n\left(\frac{z-c}{d}\right). $$