I'd like to prove that given $U,V \in \mathbb{S}^n$, we have that $$||U^++V^+||_F \geq ||(U+V)^+||_F$$ where $U^+$ denotes the projection of $U$ onto $\mathbb{S}^n_+$ with respect to the Frobenius norm. I know that given the eigenvalue decomposition $U = \sum_{i=1}^n \lambda_i(U)vv_i^T$, we have that $U^+ = \sum_{i=1}^n \max(\lambda_i(U), 0)$, and that $U^+$ is also the solution to the SDP problem $\min ||X - U||_F$ subject to $X \succeq 0$ (i.e. $U^+$ is the closest point to $U$ in $\mathbb{S}^n_+$).
I've tried manipulating the eigenvalue decompositions of $U^+$ and $V^+$ as well as some ideas using the triangle inequality and the fact that $||U^+ + V^+ - (U+V)||_F \geq ||(U+V)^+ - (U+V)||_F$, but it seems like I have had no luck with either approach. Any help is appreciated.