Consider an arbitrary $n\times n$ symmetric tridiagonal matrix $M_n$ with elements
$$(M_n)_{ij} = \begin{cases} 0, \quad i=j \\ u_{\text{min(i,j)}}, \quad |i-j|=1 \\ 0, \quad \text{else} \end{cases}$$
with all subdiagonal elements $0 \leq u_i \leq 1$. I would like to find the conditions on $u_{i}$ such that the smallest eigenvalue $\lambda_{\text{min}} \geq -1$. I know that the spectrum is symmetrical about zero, and using the Gershgorin circle theorem, we can bound the eigenvalues to be within the interval $[-\max_{k} (u_k + u_{k+1}),\max_{k} (u_k + u_{k+1})] $.
From numerical tests, we seem to have the condition that $$\lambda_{\text{min}} \geq -1 \implies \max_{k} (u_k^2 + u_{k+1}^2) \leq 1.$$ Is there a way to prove this? The closest I can find is Golub's theorem from which we know that at least one eigenvalue lies in $\left[-\sqrt{u_k^2 + u_{k+1}^2}, \sqrt{u_k^2 + u_{k+1}^2}\right], k=1,\ldots,n-2$.
Edit: From numerical diagonalization, we can make the conjecture that $$ \max_{k} (u_k^2 + u_{k+1}^2) \leq \min\{2 , \lambda_{\text{min}}^2 \} = \min\{2, (1-(\lambda_{\text{min}}+1))^2 \}$$ where the upper bound of $2$ comes from the Gershgorin theorem. Below is the numerical result. If this is true, then the above condition simply follows from it.
Let $$v_k={\sin ( k\alpha_n) \over \sin \,\alpha_n }, \qquad \alpha_n ={\pi\over n+1}, k=0,1,\ldots, n+1$$ Then $v_0=v_{n+1}=0$ and $$v_{k-1}+v_{k+1}=(2\cos \,\alpha_n) \,v_k\qquad k=1,2,\ldots, n$$ Thus the element $(v_1,v_2,\ldots,v_n)$ is the eigenvector of the matrix $M_n$ with $u_k\equiv u$ corresponding to the eigenvalue $2u\,\cos\, \alpha_n.$ We have $$ 2u\,\cos \,\alpha_n\approx 2u,\qquad n\to \infty $$ Therefore the condition $$u_k^2+u_{k+1}^2\le 1\qquad (*)$$ is not sufficient for guarranteeing that the eigenvalues of $M_n$ are located in $[-1,1].$
Clearly the condition $$u_k+u_{k+1}\le 1,\qquad k=1,2,\ldots, n-2$$ is sufficient.
The condition $(*)$ is necessary as the euclidean norms of the columns of $M_n$ must be less or equal $1$ which means that $$\lambda_{\min}\ge -1\implies u_k^2+u_{k+1}^2\le 1$$ Indeed the $$|\lambda_{\min}|=\lambda_{\max}=\|M_n\|$$ On the other hand $$\|M_n\|\ge \max (u_k^2+u_{k+1}^2)^{1/2}$$ Hence $$\lambda_{\min}^2\ge \max (u_k^2+u_{k+1}^2)$$