Using the first mean value theorem of integral
\int_0^ 1 |f'(x)| dx
= |f'(b)|
& lt= |f'(b)-f'(a)| + |f'(a)|
= |\int_a^b f''(x) dx| + k
& lt= |\int_a^b |f''(x)| dx| + k
& lt= \int_0^ 1 |f''(x)| dx + k
1. If k=0, the conclusion holds.
2. If k>0, then f(x) is strictly monotonic, and there is at most one zero point f(c)=0 on [0, 1].
2. 1) If zero c does exist, | f (x) | = | f (x)-f (c) | > =k|x-c|, get the integral.
\int_0^ 1 | f(x)| dx & gt； = k/4
such
\int_0^ 1 | f '(x)| dx & lt； = 4*\int_0^ 1 | f(x)| dx+\int_0^ 1 | f ' '(x)| dx
2.2) If f has no zero, then | f (0) | < |f( 1)|
\int_0^ 1 |f(x)| dx
= |\int_0^ 1 f(x) dx|
& gt= |\int_0^ 1 f(x)-f(0)dx+f(0)|
= |\int_0^ 1 f(x)-f(0)dx |+| f(0)|
= \int_0^ 1 | f(x)-f(0)| dx+| f(0)|
& gt= k/4 + |f(0)|
& gtk/4
There are also
\int_0^ 1 | f '(x)| dx & lt； 4*\int_0^ 1 | f(x)| dx+\int_0^ 1 | f ' '(x)| dx
Both cases are better than the inequality you want to prove.