Prove that: $$ |\prod_{i=0}^n(x-x_i)|\leq\frac{n!}{4}(\frac{b-a}{n})^{n+1} $$ where $x_i=a+i\frac{b-a}{n}$ for $i=0,...,n$ and $x\in [a;b]$
I have tried to do induction: I proved it for $n=1$ and now I need to do prove the second step of induction - proof for $n=k+1$
I found out that $x_{k+1}=b$, and I divided both sided of inequality by $|x-x_{k+1}|=b-x$ (in case $x=x_{k+1}$, the proof is trivial)
and now I basically just need to prove, that: $$ \frac{b-a}{b-x}\cdot (\frac{b-a}{k+1})^{k+1}\geq (\frac{b-a}{k})^{k+1} $$ any tips on how to do it?
The conjecture is not true:
We can simplify it to
$$\frac{b-a}{b-x} \ge \left( \frac{k+1}{k}\right)^{k+1}$$
If $k=1$, then the RHS is $4$.
Now pick $b=1, a=0$, and $x=\frac12$, we have LHS$=2$ which is a contradiction. In fact, we can choose $x$ such that the LHS gets arbitrarily close to $1$.
Your original question has been answerd here.