In the Cholesky decomposition, the argument of the square root is always positive if the matrix is real and positive definite. Why? Computing the Cholesky decomposition for an $n \times n$ matrix $A$ you need to evaluate
$$l_{jj} = \sqrt{a_{jj}-\sum_{k=1}^{j-1} l^2_{jk}}$$
The argument of the square root is always positive if $A$ is real and positive definite.  Why is that the case?
 A: Let's prove this fact by induction.
Since $A$ is positive definite then $a_{11} > 0$. It follows from the fact that
$a_{11} = \mathbf e_1^\top A \mathbf e_1 > 0$. Here $\mathbf e_1$ stands for the first column of the identity matrix. This is the base case $r = 1$ of the induction.
Suppose that we have successfully decomposed the upperleft $r \times r$ block of the matrix $A$ and the $a_{jj}-\sum_{k=1}^{j-1} l_{jk}^2$ was always positive for $j=1,\dots,r$. The relation between $l_{ij}$ and $a_{ij}$ is given by
$$
l_{jj} = \sqrt{a_{jj} - \sum_{k=1}^{j-1} l_{jk}^2}, \quad j = 1,\dots,r\\
l_{ij} = \frac{1}{l_{jj}} \left(a_{ij} - \sum_{k=1}^{j-1} l_{ik} l_{jk}\right), \quad j=1,\dots,r-1;\; i=j+1, \dots, r.
$$
Under the assumption we can safely square the first relation and multiply the second with $l_{jj} > 0$:
$$
l_{jj}^2 = a_{jj} - \sum_{k=1}^{j-1} l_{jk}^2 \Leftrightarrow a_{jj} = \sum_{k=1}^{j} l_{jk}^2\\
l_{ij}l_{jj} = a_{ij} - \sum_{k=1}^{j-1} l_{ik} l_{jk} \Leftrightarrow a_{ij} = \sum_{k=1}^{j} l_{ik} l_{jk}.
$$
There relations represent matrix equality
$$
A^{(r)} = L^{(r)} \left[L^{(r)}\right]^\top.
$$
Here $(r)$ marks upper left submatrix of $r \times r$ size. In other words leading $r\times r$ submatrix of $L$ is Cholesky decomposition of the leading $r\times r$  submatrix of $A$.
Let's start by naming different parts of $A^{(r+1)}$ and $L^{(r+1)}$:
$$
A^{(r+1)} = \begin{pmatrix}
A^{(r)} & \mathbf v\\
\mathbf v^\top & w
\end{pmatrix}\\
L^{(r+1)} = \begin{pmatrix}
L^{(r)} & \mathbf 0\\
\mathbf t^\top & \ast
\end{pmatrix}\\
$$
I've marked the $l_{r+1,r+1}$ as $\ast$ since its existence was not proven yet. On the other hand $l_{r+1,k}, k=1,\dots,r$ are well defined by
$$
l_{r+1, k} = \frac{1}{l_{k,k}} \left(
a_{r+1, k} - \sum_{m=1}^{k-1} l_{r+1, m} l_{k, m}
\right)
$$
Rewriting the same using vectors $\mathbf v$ and $\mathbf t$ gives
$$
t_k = \frac{1}{l_{k,k}} \left(
v_k - \sum_{m=1}^{k-1} t_m l_{km}
\right) \Leftrightarrow v_k = \sum_{m=1}^{k-1} t_m l_{km} + t_k l_{kk} = 
\sum_{m=1}^k t_m l_{km}.
$$
The latter is elementwise form of $\mathbf v = L^{(r)} \mathbf t$. Let's now express $a_{r+1,r+1} - \sum_{k=1}^r l_{r+1,k}^2$:
$$
a_{r+1,r+1} - \sum_{k=1}^r l_{r+1,k}^2 = w - \sum_{k=1}^r t_k^2 = w - (\mathbf t, \mathbf t).
$$
Since $A$ is positive defined for any vector $\mathbf z$ of length $r$
$$
0 < Q(\mathbf z) \equiv \begin{pmatrix}
\mathbf z^\top & 1
\end{pmatrix}
\begin{pmatrix}
A^{(r)} & \mathbf v\\
\mathbf v^\top & w
\end{pmatrix}
\begin{pmatrix}
\mathbf z \\ 1
\end{pmatrix} = \mathbf z^\top A^{(r)} \mathbf z + 2 (\mathbf v, \mathbf z) + w.
$$
I would like find such $\mathbf z$ so $Q(\mathbf z)$ is exactly $w - (\mathbf t, \mathbf t)$. Let's rewrite $Q(\mathbf z)$ as
$$
Q(\mathbf z) = \mathbf z^\top A^{(r)} \mathbf z + 2 (\mathbf v, \mathbf z) + w = 
\mathbf z^\top L^{(r)} \left[L^{(r)}\right]^\top \mathbf z + 2 (L^{(r)} \mathbf t, \mathbf z) + w = \\ =
\mathbf z^\top L^{(r)} \left[L^{(r)}\right]^\top \mathbf z + 2 (\mathbf t, \left[L^{(r)} \right]^\top\mathbf z) + w.
$$
The matrix $L^{(r)}$ is a triangular matrix with positive diagonal. Its determinant is nonzero since it is equal to product of the diagonal entries. Thus $L^{(r)}$ is invertible and we can take $\mathbf z = -\left[L^{(r)}\right]^{-\top} \mathbf t$ (the solution of $\left[L^{(r)}\right]^\top \mathbf z = -\mathbf t$). Plugging that $\mathbf z$ into $Q(\mathbf z)$ gives
$$
Q\left(-\left[L^{(r)}\right]^{-\top} \mathbf t\right) = (\mathbf t, \mathbf t) - 2 (\mathbf t, \mathbf t) + w = w - (\mathbf t, \mathbf t).
$$
Since $A$ is positive definite we know that $Q(\cdot) > 0$. On the other hand $w - (\mathbf t, \mathbf t)$ is exactly the expression under the root in the definition of $l_{r+1, r+1}$.
This finishes proof for the induction step.
