Let $A\in M_{n\times n}(\mathbb{R})$ be a real symmetric matrix and $H\in M_{n\times n}(\mathbb{C})$ be an invertible hermitian matrix.
We are given that $A+H$ and $A-H$ are positive semi-definite. Since $2A = (A+H)+(A-H)$, we can conclude that $A$ is positive semi-definite. In addition, does the invertibility of $H$ also imply that $A$ is invertible? In other words, can we conclude from the given information that $A$ is positive definite?
Take two PSD matrices, $X$ and $Y.$ The only way $X+Y$ is not definite is if $X$ and $Y$ have intersecting kernels (so, $X v = Y v = 0.$) but since $A+H - (A-H) = 2 H$ that is not possible here.