A friend has submitted this problem to me:
Let $0<a<b<1$ and $f:[0,1]\to \mathbb R$ be a differentiable function such that $$\displaystyle \frac{\int_0^a f(x) dx}{a(1-a)}+\frac{\int_b^1 f(x) dx}{b(1-b)}=0$$
Prove that $\displaystyle \left| \int_0^1 f(x)dx\right|\leq \left(\frac{b-a}2\right)\sup_{x\in [0,1]}\left|f'(x)+2\normalsize{\frac{\int_0^a f(t) dt}{a(1-a)}}\right|$
I haven't solved this yet, but I've made some progress:
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Since $\frac{1}{1-a}\left(\frac{1}{a}\int_0^a f(x) dx\right)+ \frac{1}{b}\left(\frac{1}{1-b}\int_b^1 f(x) dx\right)=0$, the Mean Value Theorem yields the existence of $\xi_a\in[0,a]$ and $\xi_b\in[b,1]$ such that $$bf(\xi_a)+(1-a)f(\xi_b)=0$$
This implies $f(\xi_a)$ and $f(\xi_b)$ have opposite signs, thus there is some $c\in [\xi_a,\xi_b]$ such that $f(c)=0$. -
We may suppose WLOG that $\sup_{x\in [0,1]} |f'(x)|<\infty$ (there's nothing to prove otherwise). Let $M=\sup_{x\in [0,1]} |f'(x)|$.
One can write $$\displaystyle \begin{align}\int_0^1 f(x)dx &= \int_0^a f(x)dx + \int_a^b f(x)dx + \int_b^1 f(x)dx \\
&= \frac{\int_0^a f(x) dx}{a(1-a)} \left( a(1-a) -b(1-b)\right) + \color{red}{\int_a^b f(x)dx} \end{align} $$
Note that $\displaystyle \color{red}{\int_a^b f(x)dx} = \color{green}{\int_a^b (f(x)-f(a))dx} + \color{blue}{(b-a)f(a)}$.
Since $f(a) = f(a)-f(c) = f'(\xi_c) (a-c)$, we have $|f(a)|\leq M |a-c|\leq M$.
The Mean Value Theorem also gives $\displaystyle \left| \int_a^b (f(x)-f(a))dx \right| \leq M \int_a^b (x-a) dx = M \frac{(b-a)^2}2$.
Putting everything together we have the estimate
$$ \left| \int_0^1 f(x)dx\right|\leq \left(\frac{b-a}2\right) \left[2 \left|\frac{\int_0^a f(x) dx}{a(1-a)}\right| \underbrace{\frac{ \left( a(1-a) -b(1-b)\right)}{b-a}}_{\leq 1} + \color{green}{M (b-a)} +\color{blue}{2M } \right]$$
and we get the bound $$ \left| \int_0^1 f(x)dx\right|\leq \left(\frac{b-a}2\right)\left(2\normalsize{\frac{\int_0^a f(t) dt}{a(1-a)}} + 3M\right) $$
which is not as sharp as what's required…
Note: this problem is similar to this Prove an integral inequality $|\int\limits_0^1f(x)dx|\leq\frac{1-a+b}{4}M$
I've tried to apply similar techniques, to no avail.
Best Answer
Let $$\lambda = \frac{\int_0^a f(x)\mathrm{d}x}{a(1-a)}= -\frac{\int_b^1 f(x)\mathrm{d}x}{b(1-b)}.$$ Set $g(x) = f(x)+2\lambda x - \lambda$. Notice that $$\int_0^a g(x)\mathrm{d}x = \int_0^a f(x)\mathrm{d} x + \lambda a^2 - \lambda a = \lambda a (1-a) + \lambda a ^2 - \lambda a = 0.$$ Similarly, we obtain $$\int_b^1 g(x)\mathrm{d}x = 0.$$
Now the left hand side of inequality becomes $$\left|\int_0^1 \left(g(x) - 2\lambda x + \lambda\right) \mathrm{d}x\right| = \left|\int_0^1 g(x) \mathrm{d}x \right|,$$ whereas the right hand side becomes $$\frac{b-a}{2}\cdot \sup_{0 \le x\le 1}|g'(x)|.$$
Indeed, we have reformulated the problem in terms of $g$. Note that the top answer for Prove an integral inequality $\left|\int_0^1f(x)dx\right|\le\frac{1−a+b}{4}\sup_{0\le x \le 1}|f'(x)|$ actually proves a stronger statement, and it is exactly what we need.