Here is the original question. The $(i, j)$-entry of the matrix $A$ is denoted by $[A]_{i,j}$.
Suppose that $A$ is an $n \times n$ matrix with real entries.
(1) If the inequality $$\begin{align*} [A]_{i,i} > \sum_{\substack{1 \leq u \leq n \\ u \neq i}} {|[A]_{i,u}|}. \end{align*}$$ holds for every positive integer $i$ less than or equal to $n$, then $\det {(A)} > 0$.
(2) If the inequality $$ \begin{align*} [A]_{j,j}\,[A]_{k,k} > \Bigg( \sum_{\substack{1 \leq p \leq n \\ p \neq j}} {|[A]_{j,p}|} \Bigg) \Bigg( \sum_{\substack{1 \leq q \leq n \\ q \neq k}} {|[A]_{k,q}|} \Bigg). \end{align*}$$ holds for every pair of distinct positive integers $j$, $k$ less than or equal to $n$, and $[A]_{i,i} > 0$ for every positive interger $i$ less than or equal to $n$, then $\det {(A)} > 0$.
I can solve the first part. Here is my proof.
Let $P(n)$ be the proposition:
The determinant of $A$ is greater than zero for every $n \times n$ matrix $A$ with real entries and with the property that the inequality $$\begin{align*} [A]_{i,i} > \sum_{\substack{1 \leq u \leq n \\ u \neq i}} {|[A]_{i,u}|}. \end{align*}$$ holds for every positive integer $i$ less than or equal to $n$.
We prove that $P(n)$ is true for every positive integer $n$ by mathematical induction.
$P(1)$ is obviously true.
We show that $P(2)$ is true. Let $$ \begin{align*} A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \end{align*} $$ be a $2 \times 2$ matrix with real entries such that $$ \begin{align*} a > |b| \quad \text{and} \quad d > |c|. \end{align*} $$ Then $$ \begin{align*} \det {(A)} = ad - bc > |b|\,|c| - bc = |bc| - bc \geq 0. \end{align*} $$
Assume that $P(n-1)$ is true. We have to show that $P(n)$ is true.
Let $A$ be an $n \times n$ matrix with real entries. Suppose that the inequality $$ \begin{align*} [A]_{i,i} > \sum_{\substack{1 \leq u \leq n \\ u \neq i}} {|[A]_{i,u}|}. \end{align*} $$ holds for every positive integer $i$ less than or equal to $n$.
Note that $[A]_{1,1}$ is certainly nonzero, since it is greater than some nonnegative number.
Let $B$ be the $n \times n$ matrix obtained from $A$ by adding $-[A]_{i,1}/[A]_{1,1}$ times row $1$ to row $i$ for $i = 2$, $3$, $\dots$, $n$. Then $$ \begin{align*} [B]_{1,j} & = [A]_{1,j}; \\ [B]_{i,1} & = 0, & \quad i > 1; \\ [B]_{i,j} & = [A]_{i,j} - \frac{[A]_{i,1}}{[A]_{1,1}} [A]_{1,j}, & \quad i > 1; \\ \det {(B)} & = \det {(A)}. \end{align*} $$
Let $C$ be the $(n-1) \times (n-1)$ matrix obtained from $B$ by removing the first row and the first column of $B$. Then $$ \begin{align*} [C]_{i,j} & = [B]_{i+1,j+1} = [A]_{i+1,j+1} - \frac{[A]_{i+1,1}}{[A]_{1,1}} [A]_{1,j+1}; \\ \det {(A)} & = \det {(B)} = [A]_{1,1} \det {(C)}. \end{align*} $$
For $j \neq k$, $$ \begin{align*} [A]_{k,k} > \sum_{\substack{1 \leq u \leq n \\ u \neq k}} {|[A]_{k,u}|} \geq |[A]_{k,j}|, \end{align*} $$ so $$ \begin{align*} [C]_{i,i} = {} & [A]_{i+1,i+1} - \frac{[A]_{i+1,1}}{[A]_{1,1}} [A]_{1,i+1} \\ = {} & \frac{[A]_{i+1,i+1} [A]_{1,1} - [A]_{1,i+1} [A]_{i+1,1}} {[A]_{1,1}} \\ > {} & \frac{|[A]_{1,i+1}|\, |[A]_{i+1,1}| - [A]_{1,i+1} [A]_{i+1,1}} {[A]_{1,1}} \\ = {} & \frac{|{[A]_{1,i+1} [A]_{i+1,1}}| - [A]_{1,i+1} [A]_{i+1,1}} {[A]_{1,1}} \\ \geq {} & 0. \end{align*} $$ Hence $0 < [C]_{i,i} = |[C]_{i,i}|$.
Note that $$ \begin{align*} |[C]_{i,i}| = {} & \Bigg| [A]_{i+1,i+1} - \frac{[A]_{i+1,1}}{[A]_{1,1}} [A]_{1,i+1} \Bigg| \\ \geq {} & |[A]_{i+1,i+1}| - \Bigg| \frac{[A]_{i+1,1}}{[A]_{1,1}} [A]_{1,i+1} \Bigg| \\ = {} & |[A]_{i+1,i+1}| - \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} |[A]_{1,i+1}|, \end{align*} $$ and that $$ \begin{align*} & \sum_{\substack{1 \leq v \leq n-1 \\ v \neq i}} {|[C]_{i,v}|} \\ = {} & \sum_{\substack{2 \leq \ell \leq n \\ \ell \neq i+1}} {|[C]_{i,\ell-1}|} \\ = {} & \sum_{\substack{2 \leq \ell \leq n \\ \ell \neq i+1}} {\Bigg| [A]_{i+1,\ell} - \frac{[A]_{i+1,1}}{[A]_{1,1}} [A]_{1,\ell} \Bigg|} \\ \leq {} & \sum_{\substack{2 \leq \ell \leq n \\ \ell \neq i+1}} {\Bigg( |[A]_{i+1,\ell}| + \Bigg| \frac{[A]_{i+1,1}}{[A]_{1,1}} [A]_{1,\ell} \Bigg| \Bigg)} \\ = {} & \sum_{\substack{2 \leq \ell \leq n \\ \ell \neq i+1}} {\Bigg( |[A]_{i+1,\ell}| + \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} |[A]_{1,\ell}| \Bigg)} \\ = {} & \sum_{\substack{2 \leq \ell \leq n \\ \ell \neq i+1}} { |[A]_{i+1,\ell}| } + \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} \sum_{\substack{2 \leq \ell \leq n \\ \ell \neq i+1}} { |[A]_{1,\ell}| } \\ = {} & \sum_{\substack{1 \leq \ell \leq n \\ \ell \neq i+1}} { |[A]_{i+1,\ell}| } - |[A]_{i+1,1}| + \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} \sum_{2 \leq \ell \leq n} { |[A]_{1,\ell}| } % \\ % {} & % \hphantom{=} - \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} |[A]_{1,i+1}| \\ = {} & \sum_{\substack{1 \leq \ell \leq n \\ \ell \neq i+1}} { |[A]_{i+1,\ell}| } - \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} |[A]_{1,1}| + \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} \sum_{2 \leq \ell \leq n} { |[A]_{1,\ell}| } % \\ % {} & % \hphantom{=} - \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} |[A]_{1,i+1}| \\ = {} & \sum_{\substack{1 \leq \ell \leq n \\ \ell \neq i+1}} { |[A]_{i+1,\ell}| } - \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} \Bigg( |[A]_{1,1}| - \sum_{2 \leq \ell \leq n} { |[A]_{1,\ell}| } \Bigg) % \\ % {} & % \hphantom{=} - \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} |[A]_{1,i+1}| \\ \leq {} & \sum_{\substack{1 \leq \ell \leq n \\ \ell \neq i+1}} { |[A]_{i+1,\ell}| } - \frac{|[A]_{i+1,1}|}{|[A]_{1,1}|} |[A]_{1,i+1}|. \end{align*} $$ Hence $$ \begin{align*} |[C]_{i,i}| - \sum_{\substack{1 \leq v \leq n-1 \\ v \neq i}} {|[C]_{i,v}|} \geq |[A]_{i+1,i+1}| - \sum_{\substack{1 \leq \ell \leq n \\ \ell \neq i+1}} { |[A]_{i+1,\ell}| } > 0. \end{align*} $$
Thus by assumption, $\det {(C)} > 0$.
Because $[A]_{1,1} > 0$ and $\det {(C)} > 0$, we conclude that $\det {(A)} = [A]_{1,1} \det {(C)} > 0$.
We have shown that $P(n)$ is true.
By mathematical induction, $P(n)$ is true for $n = 1$, $2$, $3$, $\dots$.
However, I am not able to solve the second part. Mathematical induction does not seem to work. Perhaps it should be solved differently, but I have no idea.
Thanks for the hints in the comments.
We note that (2) implies (1), so it suffices to prove (2).
We do it with Brauer's oval theorem. Suppose that an $n \times n$ matrix $A$ with real entries meets the requirements in (2). Let $\lambda = x + \mathrm{i} y$ (in which $x$, $y \in \mathbb{R}$) be an eigenvalue of $A$. Put $$ \begin{align*} R_i = \sum_{\substack{1 \leq u \leq n \\ u \neq i}} {|[A]_{i,u}|}. \end{align*} $$ We claim that $x > 0$, from which we can conclude that $\det {(A)} > 0$. If $x \leq 0$, it would be true by Brauer's oval theorem that $$ \begin{align*} R_j R_k \geq {} & |{\lambda - [A]_{j,j}}|\,|{\lambda - [A]_{k,k}}| \\ = {} & |{[A]_{j,j} - \lambda}|\,|{[A]_{k,k} - \lambda}| \\ = {} & |{([A]_{j,j} - x) - \mathrm{i}y}|\, |{([A]_{k,k} - x) + \mathrm{i}y}| \\ \geq {} & |{[A]_{j,j} - x}|\,|{[A]_{k,k} - x}| \\ = {} & ([A]_{j,j} - x)\,([A]_{k,k} - x) \\ \geq {} & [A]_{j,j}\,[A]_{k,k} \\ > {} & R_j R_k \end{align*} $$ for some distinct integers $j$, $k$. A contradiction.
My old proof of (1) is just brute force. It gives me little insight about (2).