Conditions ensuring an order betweenthe  smallest eigenvalues of two positive definite Jacobi matrices Let $J, L$ be two symmetric positive definite tridiagonal matrices of positive diagonal entries,  $\mbox{diag}(J)=(a_1, a_2, \ldots, a_n)$, $\mbox{diag}(L)=(\alpha_1, \alpha_2, \ldots, \alpha_n)$, with  $\min_{i=1}^n(a_i)<\min_{i=1}^n(\alpha_i)$ and $\mbox{trace}(L-J)>0$. The two matrices have constant off diagonal entries, $\mbox{offdiag}(J)=\mbox{offdiag}(L)=(b,b,b,\ldots, b)$, $b<0$.
I'm interested whether or not these conditions are sufficient to ensure that the smallest eigenvalue of $J$ is less than the smallest eigenvalue of $L$.
 A: No, your conditions are not sufficient to ensure that the smallest eigenvalue of $J$ is less than the smallest eigenvalue of $L$. Here is a simple counterexample,
$$
\begin{equation*}
 J = 
 \begin{bmatrix}
 1.9  & -1 &   0\\\\
 -1 & 4  &  -1\\\\
     0   &-1  &  7
\end{bmatrix},\qquad
 L = 
  \begin{bmatrix}
   2 &  -1 &  0\\\\
   -1 & 3  & -1\\\\
         0 &  -1 & 10
 \end{bmatrix}.
\end{equation*}
$$
Here $1.9 = \min(\mbox{diag}(J)) < \min(\mbox{diag}(L))=2$, $\mbox{trace}(L-J) = 2.1$, but $\lambda_{\min}(J) = 1.4736$ while $\lambda_{\min}(L) = 1.3488$.

Older, messier counterexample.
$$
\begin{equation*}
 J = \begin{bmatrix}
 0.3309  & -0.0463 &   0\\\\
 -0.0463 & 0.5364  &  -0.0463\\\\
     0   &-0.0463  &  0.5951
\end{bmatrix},\qquad
 L = \begin{bmatrix}
   0.3432 &  -0.0463 &  0\\\\
   -0.0463 & 0.4117  & -0.0463\\\\
         0 &  -0.0463 & 0.7940
 \end{bmatrix}.
\end{equation*}
$$
Here we see that the $0.3309 = \min(\mbox{diag}(J)) < \min(\mbox{diag}(L))=0.3432$, $\mbox{trace}(L-J) = 0.0865$, but $\lambda_{\min}(J) = 0.3206$ while $\lambda_{\min}(L) = 0.3190$.
A: Well, very roughly speaking, if $b$ is small relative to the diagonals, then the eigenvalues of $J,L$ will be approximately their diagonal values and your conclusion will hold.
A: Felix is correct in the small $b$ limit. In the very large $b$ limit, on the other hand, the spectra of $J$ and $L$ will coincide, so one needs some more finesse. The following is a rough, approximate attempt at making some intuition.
Consider the case where the first diagonal entry of $J$ is $a_1<0$ with $|a_1|\ll |b|^2$, and all other diagonal entries in $J$ and $L$ are zero. Then you can expand $\det(tI-J)$ along the first row to get $$\det(tI-J)=b^{2n} \left(U_n(t/2)-\frac{a_1 }{b^2} U_{n-1}(t/2)\right),$$
where $U_n$ is the $n$th Chebyshev polynomial of the second kind. For small $a_1$, then, the question is which way does the second term "push" the eigenvalue? I don't have a proof but it's graphically clear that the signs of $U_n$ and $U_{n-1}$ coincide just inside of the former's leftmost zero, which means that a small, negative $a_1$ will push $J$'s minimum eigenvalue left, in accord with your conjecture.
This can be extended to $L$ having a nonzero element in its diagonal, which will push its minimum eigenvalue either left or right but no more than $J$'s, at least at first order.
Since your result is (or appears to be) true both for big and small $b$, I should expect it to be true everywhere, or to have some pretty interesting mathematics in the middle.
