Consider the matrix $2\times2$ symmetric matrix:
$$
A\_2=\begin{pmatrix} 1 & a\_1 \\ a\_1 & 1\end{pmatrix}.
$$
It's clear that the restriction $|a\_1|<1$ implies that $\det(A\_2)>0$. Moreover, this is the best restriction on the modulus of $a\_1$ with this property, for $\det\begin{pmatrix} 1 & 1\\ 1 & 1\end{pmatrix}=0$.


Now, consider the $3\times3$ symmetric matrix:
$$
A\_3= \begin{pmatrix} 1 & a\_1 & a\_2\\ a\_1 & 1 & a\_3 \\ a\_2 & a\_3 & 1\end{pmatrix}
$$
We have $\det(A\_3)=1+2a\_1a\_2a\_3-a\_1^2-a\_2^2-a\_3^2$. Then, the restriction $|a\_i|<1/2$ implies that
$$
\det(A\_3)>1-2(1/8)-3(1/4)=0.
$$
Moreover, this is the best restriction on the modulus of the $a\_i$'s with this property, for
$$
\det\begin{pmatrix} 1 & -1/2 & -1/2\\ -1/2 & 1 & -1/2 \\ -1/2 & -1/2 & 1\end{pmatrix}=0.
$$


Now we consider the general case. Let $A\_n=[a\_{i,j}]\_{1\leq 1,j\leq n}$ be a matrix such that $a\_{i,j}=a\_{j,i}$ for every $1\leq i,j\leq n$ and $a\_{i,i}=1$ for every $1\leq i\leq n$. What is the greatest number $\alpha\_n$ satisfying
$$
\max\_{i\neq j}|a\_{i,j}| < \alpha\_n\ \ \Rightarrow\ \ \det A\_n>0?
$$


I was able to prove that $\alpha\_n\leq 1/(n-1)$, by showing that the determinant of the following $n\times n$ matrix is $0$:
$$
M\_n=\begin{pmatrix} 1 & -\dfrac{1}{n-1} & -\dfrac{1}{n-1} & \dots & -\dfrac{1}{n-1} \\ 
-\dfrac{1}{n-1} & 1 & -\dfrac{1}{n-1} & \cdots & -\dfrac{1}{n-1} \\
-\dfrac{1}{n-1} &-\dfrac{1}{n-1}& 1 &\dots & -\dfrac{1}{n-1} \\
\vdots & \vdots & \vdots &\ddots & \vdots \\
-\dfrac{1}{n-1} & -\dfrac{1}{n-1}& -\dfrac{1}{n-1}&\dots & 1\end{pmatrix}.
$$


One way to see that $\det(M\_n)=0$ is to pick $v\_1,...,v\_n\in\mathbb R^n$ such that $\|v\_i\|=1$ for every $i\in\{1,...,n\}$ and $\{v\_1,...,v\_n\}$ is the set of vertices of a regular $(n-1)$-symplex. One may se by induction that $\langle v\_i, v\_j \rangle=-1/(n-1)$ for every $i\neq j$. Therefore,
$$
M\_n=\left[\langle v\_i, v\_j \rangle\right]\_{1\leq i, j \leq n}.
$$
Since $v\_1,...,v\_n$ are the vertices of an $(n-1)$-symplex, they are linearly dependent, so 
$$
\det\left(\left[\langle v\_i, v\_j \rangle\right]\_{1\leq i, j \leq n}\right)=0.
$$


My intuition says that $\alpha\_n=1/(n-1)$. However, I'm not able to prove that $\alpha\_n \geq 1/(n-1)$. Any suggestions would be very helpful. Thanks in advice.



On the determinant of a class symmetric matrices
Your guess is correct. If the elements outside the diagonal have absolute values less than $1/(n-1)$, the matrix has 'diagonal dominance', thus it is nonsingular. 


To make the answer self-contained, I give a proof. If $x=(x\_1,\dots,x\_n)^t$ satisfies $Ax=0$, take $k$ such that $|x\_k|$ is maximal and look at $\sum a\_{ki}x\_i$. The summand $a\_{kk}x\_k$ has greater absolute value than all other summands altogether, thus the total sum is non-zero. A contradiction. 


So, determinant of $A$ is non-zero. And it is non-zero if we multiply all elements outside diagonal by $t\in [0,1]$, call such a new matrix $A(t)$. Since $f(t)=\det A(t)$ is continuous, does not vanish on $[0,1]$ and $f(0)=1$, we get $f(1)>0$ as desired.


