Is there a Schur-like theorem for proving $B^TB\succeq A$?

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This is in the context of semidefinite programs, and $A\in\mathbb R^{n\times n},~B\in\mathbb R^{k\times n}$ are variables. If want to show that $A-B^TB\succeq 0$ and I know $A\succeq 0$, then I can instead show that $X=\begin{bmatrix} A&B\\ B^T&I \end{bmatrix}\succeq 0$. This is nice because it is convex.

On the other hand, I'd like to find a convex representation for $B^TB-A\succeq 0$, given that $A\succeq 0$. Of course, we can't use Schur's complement directly to solve this, but I'm wondering if there's a technique to represent this as a convex constraint or if there's no way in general?