If I'm right the Wirtinger's inequality (this section of Wikipedia) implies $$\int_0^1\left|\sum_{n=1}^\infty\frac{\mu(n)}{n^2}\sin(n\pi x)\right|^2dx\leq\int_0^1\left|\sum_{n=1}^\infty\frac{\mu(n)}{n}\cos(n\pi x)\right|^2dx,\tag{1}$$ where for integers $n\geq 1$ we are denoting the Möbius function as $\mu(n)$ (you can see the definition of this arithmetic function, for example from Wikipedia or MathWorld).
Question. I would like to know if it is possible to state an estimation of such difference in $(1)$, say us $$\left|\int_0^1\left(\left|\sum_{n=1}^\infty\frac{\mu(n)}{n}\cos(n\pi x)\right|^2-\left|\sum_{n=1}^\infty\frac{\mu(n)}{n^2}\sin(n\pi x)\right|^2\right)dx\right|.\tag{2}$$ Many thanks.
Thus I am asking about what work can be done to know something about the size of such difference of integrals (I wrote the absolute value in $(2)$, but I'm interested about what work can be done about the difference of the sides of $(1)$).
Both terms can be computed in a explicit way by Parseval's identity. $$\int_{0}^{1}\left(\sum_{n\geq 1}\frac{\mu(n)}{n^2}\sin(\pi n x)\right)^2\,dx =\frac{1}{2}\sum_{n\geq 1}\frac{\mu(n)^2}{n^4}=\frac{1}{2}\prod_{p}\left(1+\frac{1}{p^4}\right)=\frac{\zeta(4)}{2\zeta(8)}=\frac{105}{2\pi^4} $$ and similarly $$\int_{0}^{1}\left(\sum_{n\geq 1}\frac{\mu(n)}{n}\cos(\pi n x)\right)^2\,dx =\frac{1}{2}\sum_{n\geq 1}\frac{\mu(n)^2}{n^2}=\frac{1}{2}\prod_{p}\left(1+\frac{1}{p^2}\right)=\frac{\zeta(2)}{2\zeta(4)}=\frac{15}{2\pi^2} $$ so you just discovered that $\pi^2\geq 7$ is a consequence of Wirtinger's inequality.
On the other hand the sharper $\pi>3$ is already a consequence of convexity: $6$ is the perimeter of a regular hexagon inscribed in the unit circle, and if $A,B$ are two convex, bounded sets in the plane such that $A\subsetneq B$, the length of $\partial A$ is less than the length of $\partial B$. And we can do much better by considering $$\frac{1}{4^4}\geq \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx = \frac{22}{7}-\pi $$ which shows that the Archimedean approximation is an upper bound for $\pi$, but $\pi > \frac{113}{36}$.