Integral inequality between a function and its derivative The problem is:
Let a function $f$ be continuously differentiable on [0,1]. Prove that
$$
\int\limits_0^1|f(x)|\,dx\le \max\left\{ \int\limits_0^1 |f'(x)| \, dx, \left| \int\limits_0^1 f(x) \, dx\right|\right\}
$$
For $f$ nonpositive or nonnegative this is obvious. Intuitively I know, that it is true in the remaining case, where the first integral of the righthand side comes into play, which is a sum of $f$ changes between local maxima and minima. Please, help to prove this inequality strictly.
 A: This is a very intriguing problem, when one sees that, as Paul Sinclair pointed out, 
$$\int_0^1|f(x)|dx \geq \left|\int_0^1f(x)dx\right|$$
This at first quite confuses us, but this insight is in fact integral to the solution. This tells us that the equality case here is important, for otherwise, having this in the maximum function would be meaningless. So, let us examine the equality case. What is important to see here is that iff
$$\int_0^1|f(x)|dx = \left|\int_0^1f(x)dx\right|$$
Then for $x \in [0,1]$, we know that the sign of $x$ is constant. Now that we know this, we know that the inequality does indeed hold as long as the sign of $x$ is constant. Next we examine the other case.
If the sign of $x$ is not constant, we can make the following argument. Let, $a$ be the maximum of f over the range $[0,1]$ and let $-b$ be the minimum of f over the same range. We know, because there is a sign change, that $b$ and $a$ are positive values. Next, we know that from the fundamental theorem of calculus,
$$\int_0^1 |f'(x)|dx \geq \left|\int_{f^{-1}(-b)}^{f^{-1}(a)} f'(x)dx \right| \geq f(f^{-1}(a)) - f(f^{-1}(-b)) = a + b$$
And finally, we also know that from the inequality $-b < f(x) < a$, we have
$$\int_0^1 |f(x)dx| < \max(a,b) < a + b \leq\int_0^1 |f'(x)|dx$$
In this way, the inequality holds for both the case where the sign is constant as well as the case where it isn't, and thus we are done.
