Show that the matrix, $$ A= \begin{bmatrix} 1 & \rho & \dots & \rho \\ \rho & 1 & \dots & \rho \\ \vdots & \rho & \ddots & \vdots \\ \rho & \dots & \rho & 1 \end{bmatrix} $$ is positive-definite if and only if $\frac{-1}{n-1}<\rho<1$.
I know that one way would be to show that all eigenvalues are positive but I couldn't produce anything usefuol yet. I would be glad for some ideas or rough proof sketch.
Setting aside the degenerate case $n=1$ for which $A$ is positive-definite for all $\rho$, here's a sketch of a purely mathematical solution,
Consider the decomposition of your matrix,
$$A(\rho)=\rho{\bf 11}^T + (1-\rho)I.$$
Let $A_k(\rho)$ denote the upper-left $k\times k$ submatrix of $A(\rho)$.
Then $A(\rho)$ being positive-definite is equivalent to all of its leading principal minors $\det A_k(\rho)$ for $1\leq k\leq n$ being positive.
Therefore we must show that for all $1\leq k\leq n$,
\begin{align} \det A_k(\rho)&=\det (\rho{\bf 11}_k^T + (1-\rho)I_k)\\ &=(1-\rho)^k+k\rho(1-\rho)^{k-1} \end{align}
is positive on the open interval $(-\frac{1}{n-1},1)$, and that for any point outside this interval there exists a $k$ such that it's non-positive, where the second equality follows by the matrix determinant lemma.
Hint: take the derivative with respect to $\rho$.
