I want to prove that given a connected closed (compact without boundary) oriented Riemmanian manifold $(M,g)$, the condition $$\int_M \varphi \;\mathrm{d}\mathrm{vol}_g=0$$ implies that $\int \varphi h \;\mathrm{d}\mathrm{vol}=0$ for every harmonic function $h$, i.e., every function such that $\triangle h:=(dd^*+d^*d)h=0$.
Note: By the Hodge Decomposition Theorem, we have $C^\infty(M,\mathbb{R})=\mathrm{Ker}(\triangle) \oplus \mathrm{Im}(\triangle)$, so this would prove that the given condition implies the solution of the Poisson problem $\triangle f=\varphi$ for $\varphi$ smooth.
In fact, the condition is also necessary, but that is easier to prove.
I know this condition must be true, but seen in this way it does not seem to obvious. It is saying that if it is orthogonal to the constant function $1$, then it is orthogonal to every harmonic function.
Note 2: There's a similar question made before. But the point is that I would not like to solve the Poisson equation appealing to PDE theory and instead try to prove it via the Hodge Decomposition Theorem (the standard PDE path is to use Riesz representation theorem and then elliptic regularity, but one needs Sobolev spaces and elliptic regularity; I have the intuition that there must be a simple proof using the Hodge decomposition argument).