Problem: Suppose $f(x)$ has continuous $(n+1)$-th derivative over $[a,b]$, and $f(a)=f'(a)=\cdots=f^{(n)}(a)=0$, prove that $$ \max\limits_{a\le x\le b}|f(x)|\le\frac{(b-a)^n}{n!}\int_a^b|f^{(n+1)}(x)|\mathrm{d}x $$
My attempt: Clearly this problem has something to do with Taylor expansion. We take the Taylor expansion of $f(x)$ at $x=a$: $$ f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} $$ where $\xi\in(a,x)$.
Also, since $(b-a)^n$ appears in the formula, we consider $$ f(b)=\frac{f^{(n+1)}(\xi_b)}{(n+1)!}(b-a)^{n+1} $$ where $\xi_b\in(a,b)$.
But I don't know how to deal with $\max\limits_{a\le x\le b}|f(x)|$. I can't think of a formula containing $\max |f(x)|$. Can you help me?
You can use Taylor's theorem with the integral remainder: $$ f(x)=f(a)+f'(a)(x-a)+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n \, dt $$ so that in your case $$ |f(x)| = \left| \int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n\, dt\right| \le \frac{1}{n!}\int_a^x |f^{(n+1)}(t)| \cdot (x-t)^n\, dt \\ \le \frac{1}{n!}\int_a^b |f^{(n+1)}(t)| \cdot (b-a)^n\, dt = \frac{(b-a)^n}{n!}\int_a^x |f^{(n+1)}(t)| \, dt $$
For a proof of Taylor's theorem with the integral remainder, see for example Derivation for the integral form of the remainder, it is essentially repeated integration by parts.