Let $f$ be a differentiable function on $[0,1]$ such that $f(0)=0$ and $f(1)=1$. If the derivative $f'$ of $f$ is also continuous on $[0,1]$, prove that:
$ \int_0^1 |f'(x)-f(x)| dx \geq \frac{1}{e}$.
A hint given is to let $h(x)$ to be $e^{-x}f(x)$. I have tried to differentiate both sides to obtain $h'(x)$ = $e^{-x}(f'(x)-f(x))$, which contains the $f'(x)-f(x)$, but I don't have any idea of getting rid of the absolute value sign.
Your help will be really valuable.
The rest is quite straightforward, using the triangle inequality. Note that $$ \frac 1e = h(1) - h(0) = \int_0^1 \exp(-x)\bigl(f'(x)-f(x)\bigr)\, dx $$ Hence \begin{align*} \frac 1e &\le \int_0^1 \left|\exp(-x)\bigl(f'(x)-f(x)\bigr)\right|\, dx\\ &\le \int_0^1 \exp(-x)\left|f'(x) - f(x)\right|\, dx\\ &\le \int_0^1 \left|f'(x) - f(x)\right|\, dx \end{align*}