Suppose $f$ is continuous and that $f$ is $C^1$. Prove that $$\max_{x\in[a, b]}|f(x)|\;\leq\; \frac{1}{b-a}\Big|\int_a^b f(x)dx\Big| \,+\, \int_a^b |f'(x)|dx.$$ (Hint: how are the maximum of $f(x)$ and its average related by the fundamental theorem of calculus?)
Could somebody provide an example proof?
Observe that $$\int_{a}^{b} |f'| \geq \max_{s,t \in [a,b]} \left|\int_{t}^{s} f'\right| =\max_{s,t \in [a,b]} |f(s)-f(t)| \geq \max_{s,t \in [a,b]} |f(s)|-|f(t)| = \max_{s \in [a,b]} |f(s)| - \min_{s \in [a,b]} |f(s)|$$
and $$\frac{1}{b-a} \left|\int_{a}^{b} f\right| = |f(c)|$$ for some $c \in [a,b]$ by the mean value theorem.
The result follows from the fact that $|f(c)| \geq \min_{s \in [a,b]} |f(s)|$.