Find a continuous function, over the unit interval, satisfying $\int_0^x f(u)du\geq \sum_{n=1}^\infty\frac{\mu(n)}{n^2}x^{n-1}$ for each $ x\in[0,1]$

Question

After I've read a problem from [1] (it is a Spanish journal) I've consider the next question that seems curious to me, since now I don't know how get such simple example.

Question. We denote for integers $n\geq 1$ the Möbius function as $\mu(n)$, see the definition of this arithmetic function for example from this MathWorld. Can you provide me an (simple) example of a (real) continuous function $f$ over $[0,1]$ satisfying $$\int_0^x f(u)du\geq \sum_{n=1}^\infty\frac{\mu(n)}{n^2}x^{n-1}$$ for each $x\in[0,1]$? Many thanks.

I tried simple functions as constants, monomials and exponentials, I'm not even asking for a function such that the difference $RHS-LHS$ is small, I would like to know just an example. And it is just a curiosity. Then please, how to find it?

References:

[1] PROBLEMA 272, by Marcel Chiriţă, La Gaceta de la RSME, Vol. 19 (2016), Núm. 1, page 111.

Answer 1

No solution, as $\int_0^0 f(x){\mathrm d}x=0$, but the RHS evaluates to $1$ for $x=0$.

Answer 2

It's impossible. If f is continuous over the compact set $[0,1]$ so is bounded and since it is continuous it is Reimann integrable therefore$$\lim_{x\to 0^+}\int_{0}^{x}f(u)du=0$$also using $\epsilon-\delta$ approach we conclude that if $x<\delta=\dfrac{\epsilon}{\epsilon+1}$ therefore$$|\sum_{n=1}^\infty\frac{\mu(n)}{n^2}x^{n-1}-1|=|\sum_{n=2}^\infty\frac{\mu(n)}{n^2}x^{n-1}|\le\sum_{n=2}^\infty\frac{|\mu(n)|}{n^2}x^{n-1}\le\sum_{n=2}^\infty\frac{1}{n^2}x^{n-1}\le\sum_{n=2}^\infty x^{n-1}\le\sum_{n=2}^\infty (\dfrac{\epsilon}{\epsilon+1})^{n-1}=\dfrac{\dfrac{\epsilon}{\epsilon+1}}{1-\dfrac{\epsilon}{\epsilon+1}}=\epsilon$$which yields to $$\lim_{x\to0^+}\sum_{n=1}^\infty\frac{\mu(n)}{n^2}x^{n-1}=1$$which is a contradiction to $\lim_{x\to 0^+}\int_{0}^{x}f(u)du=0$ therefore no such $f(x)$ exists.