Positive semi-definite matrix problem If $A，B$ and $M$ are positive semi-definite matrices, and we have
$$ A+B \succeq M .$$
Do there always exist two positive semi-definite matrices $ M_{1}, M_{2}  , $ such that
$$   A \succeq M_{1},    \quad B \succeq M_{2},   \quad M_{1}+M_{2}=M  ?$$

The statement is false even when we impose stronger conditions of positive definiteness on the matrices.
Let $p > 0$ be a number to be determined. Consider the following martices:
$$
A = \begin{pmatrix}2p + 1& 0\\0&1\end{pmatrix}, \quad
B = \begin{pmatrix}1 & 0\\0 & 2p + 1\end{pmatrix} \quad\text{ and }\quad
M = \begin{pmatrix}p+1 & p\\p & p+1\end{pmatrix}
$$
It is clear $A, B, M \succ 0$ and $A + B - M = \begin{pmatrix}p+1&-p\\-p&p+1\end{pmatrix} \succ 0 \implies A + B \succ M$
If $M_1 = \begin{pmatrix}a_1&b_1\\b_1&c_1\end{pmatrix} \succeq 0$ and $A \succeq M_1$, we must have:
$$0 \le a_1 \le 2p+1,\quad 0\le c_1 \le 1\quad\text{ and }\quad a_1 c_1 - b_1^2 \ge 0$$
This implies $|b_1| \le \sqrt{a_1c_1} \le \sqrt{2p+1}$. By a similar argument, if 
$M_2 = \begin{pmatrix}a_2&b_2\\b_2&c_2\end{pmatrix} \succeq 0$ and $B \succeq M_2$, we will have $|b_2| \le \sqrt{2p+1}$. 
If we choose $p$ such that $$2\sqrt{2p+1} < p\quad\iff\quad p > 4 + 2\sqrt{5}$$
then there is no way such a pair of $M_1, M_2$ can sum to $M$. The off diagonal element simply won't match.
