[proofplan]
We prove each equivalence by writing the quadratic form of $M$ as a completed square plus the quadratic form of the corresponding Schur complement. When $A$ is positive definite, the change of variables $u \mapsto u + A^{-1}Bv$ is available because $A$ is invertible, and the square term is always non-negative. The remaining term is exactly $v^\top(C - B^\top A^{-1}B)v$, so non-negativity of $M$ for all pairs $(u,v)$ is equivalent to positive semidefiniteness of that Schur complement. The case where $C$ is positive definite is proved by the analogous completion of the square in the $v$ variable.
[/proofplan]
[step:Complete the square using the positive definite block $A$]
Assume that $A$ is positive definite. Since $A$ is symmetric and positive definite, $A$ is invertible: if $Ax = 0$ for some $x \in \mathbb{R}^p$ with $x \neq 0$, then $x^\top Ax = 0$, contradicting positive definiteness. Since $A^\top = A$, its inverse also satisfies $(A^{-1})^\top = A^{-1}$.
Define the Schur complement of $A$ in $M$ to be the symmetric matrix
\begin{align*}
S_A := C - B^\top A^{-1}B \in \mathbb{R}^{q \times q}.
\end{align*}
For $u \in \mathbb{R}^p$ and $v \in \mathbb{R}^q$, expanding the square gives
\begin{align*}
(u + A^{-1}Bv)^\top A(u + A^{-1}Bv) = u^\top Au + 2u^\top Bv + v^\top B^\top A^{-1}Bv.
\end{align*}
Therefore the quadratic form of $M$ satisfies
\begin{align*}
u^\top Au + 2u^\top Bv + v^\top Cv = (u + A^{-1}Bv)^\top A(u + A^{-1}Bv) + v^\top S_Av.
\end{align*}
[guided]
Assume that $A$ is positive definite. The first point is that the expression $A^{-1}Bv$ is legitimate. Positive definiteness implies invertibility: if $Ax = 0$ for a non-zero vector $x \in \mathbb{R}^p$, then
\begin{align*}
x^\top Ax = x^\top 0 = 0,
\end{align*}
which contradicts $x^\top Ax > 0$. Hence $A^{-1}$ exists. Since $A$ is symmetric, $A^{-1}$ is symmetric as well, because
\begin{align*}
(A^{-1})^\top = (A^\top)^{-1} = A^{-1}.
\end{align*}
Define the Schur complement of $A$ in $M$ by
\begin{align*}
S_A := C - B^\top A^{-1}B \in \mathbb{R}^{q \times q}.
\end{align*}
This matrix is symmetric because $C$ is symmetric and
\begin{align*}
(B^\top A^{-1}B)^\top = B^\top (A^{-1})^\top B = B^\top A^{-1}B.
\end{align*}
Now we complete the square in the $u$ variable. For fixed $u \in \mathbb{R}^p$ and $v \in \mathbb{R}^q$, expand the shifted quadratic term:
\begin{align*}
(u + A^{-1}Bv)^\top A(u + A^{-1}Bv) = u^\top Au + u^\top Bv + v^\top B^\top A^{-1}Au + v^\top B^\top A^{-1}Bv.
\end{align*}
Let $I_p \in \mathbb{R}^{p \times p}$ denote the identity matrix. Since $A^{-1}A = I_p$ and $v^\top B^\top u$ is the scalar transpose of $u^\top Bv$, the two mixed terms are equal:
\begin{align*}
v^\top B^\top u = u^\top Bv.
\end{align*}
Thus
\begin{align*}
(u + A^{-1}Bv)^\top A(u + A^{-1}Bv) = u^\top Au + 2u^\top Bv + v^\top B^\top A^{-1}Bv.
\end{align*}
Subtracting the last term and adding $v^\top Cv$ gives
\begin{align*}
u^\top Au + 2u^\top Bv + v^\top Cv = (u + A^{-1}Bv)^\top A(u + A^{-1}Bv) + v^\top(C - B^\top A^{-1}B)v.
\end{align*}
By the definition of $S_A$, this is
\begin{align*}
u^\top Au + 2u^\top Bv + v^\top Cv = (u + A^{-1}Bv)^\top A(u + A^{-1}Bv) + v^\top S_Av.
\end{align*}
[/guided]
[/step]
[step:Deduce the equivalence with the Schur complement of $A$]
Suppose first that $S_A \succeq 0$. For all $u \in \mathbb{R}^p$ and $v \in \mathbb{R}^q$, the completed-square identity gives
\begin{align*}
(u,v)^\top M(u,v) = (u + A^{-1}Bv)^\top A(u + A^{-1}Bv) + v^\top S_Av \geq 0,
\end{align*}
because $A \succ 0$ and $S_A \succeq 0$. Hence $M \succeq 0$.
Conversely, suppose that $M \succeq 0$. For an arbitrary $v \in \mathbb{R}^q$, set
\begin{align*}
u := -A^{-1}Bv \in \mathbb{R}^p.
\end{align*}
Then $u + A^{-1}Bv = 0$, and the completed-square identity yields
\begin{align*}
0 \leq (u,v)^\top M(u,v) = v^\top S_Av.
\end{align*}
Since this holds for every $v \in \mathbb{R}^q$, we have $S_A \succeq 0$. Therefore
\begin{align*}
M \succeq 0 \iff C - B^\top A^{-1}B \succeq 0.
\end{align*}
[/step]
[step:Complete the square using the positive definite block $C$]
Assume that $C$ is positive definite. As above, $C$ is invertible and $(C^{-1})^\top = C^{-1}$. Define the Schur complement of $C$ in $M$ to be the symmetric matrix
\begin{align*}
S_C := A - BC^{-1}B^\top \in \mathbb{R}^{p \times p}.
\end{align*}
For $u \in \mathbb{R}^p$ and $v \in \mathbb{R}^q$, expanding the square in the $v$ variable gives
\begin{align*}
(v + C^{-1}B^\top u)^\top C(v + C^{-1}B^\top u) = v^\top Cv + 2u^\top Bv + u^\top BC^{-1}B^\top u.
\end{align*}
Hence
\begin{align*}
u^\top Au + 2u^\top Bv + v^\top Cv = u^\top S_Cu + (v + C^{-1}B^\top u)^\top C(v + C^{-1}B^\top u).
\end{align*}
[/step]
[step:Deduce the equivalence with the Schur complement of $C$]
Suppose first that $S_C \succeq 0$. For every $u \in \mathbb{R}^p$ and $v \in \mathbb{R}^q$,
\begin{align*}
(u,v)^\top M(u,v) = u^\top S_Cu + (v + C^{-1}B^\top u)^\top C(v + C^{-1}B^\top u) \geq 0,
\end{align*}
because $S_C \succeq 0$ and $C \succ 0$. Thus $M \succeq 0$.
Conversely, suppose that $M \succeq 0$. For an arbitrary $u \in \mathbb{R}^p$, set
\begin{align*}
v := -C^{-1}B^\top u \in \mathbb{R}^q.
\end{align*}
Then $v + C^{-1}B^\top u = 0$, so the completed-square identity gives
\begin{align*}
0 \leq (u,v)^\top M(u,v) = u^\top S_Cu.
\end{align*}
Since this holds for every $u \in \mathbb{R}^p$, $S_C \succeq 0$. Therefore
\begin{align*}
M \succeq 0 \iff A - BC^{-1}B^\top \succeq 0.
\end{align*}
This proves both Schur complement criteria.
[/step]
