Let $f \in C_1([0,1])$
$\int |f| d\lambda \leq \int \max |f|\, d\lambda, \text{ so }\|f\|_1 \leq \|f\|_2.$
Now, since f' is continuous on closed interval $[0,1]$ it is bounded. Therefore for each $x \in [0,1]$ we derive
\begin{align} \|f\|_2 &= \max|f| + \max|f'|\newline &\leq |f(x)| + \max|f'| + \max|f'|\newline &\leq |f(x)| + 2\|f\|_1 - 2\int |f|d\lambda\newline &\leq |f(x)| + 2\|f\|_1. \end{align}
However we know that $\exists_{x_0} |f(x_0)| \leq \int |f|\, d\lambda$, so we have $|f(x_0)| \leq \|f\|_1$ and finally
$\|f_2\| \leq 3\|f\|_1.$
Why \max|f| \leq |f(x)| + \max |f'|?
Let $x,y \in [0,1]$. Then $\exists_{c \in [0,1]}$ such that \big| |f(x)| - |f(y)|\big| \leq |f(x) - f(y)| \leq |f'(c)| \leq \max |f'| by the mean value theorem.