Given this matrix

$M=\begin{bmatrix} 0 & m-1 & 2 & n-1\\ 1 & m-2& 1 & n-1\\ 2 & m-1 & 0 & 2(n-1)\\ 1 & m-1 & 2 & 2(n-2)\end{bmatrix},$ show that if $\alpha_1,\alpha_2$ are two negative eigenvalues of $M$, then $\alpha_1+\alpha_2<-3$.

Using Gershgorin circles theorem, the eigenvalues would lie in the intervals $B(0,m+n)$, $B(m-2,n-1)$, $B(0,2n+m-1)$,$ B(2n-4,m+2)$ where $B(x,y)$ denotes the interval with centre $x$ and radius $y$. But it's not helping to find the required bound?

Does there exist any theorem/trick to find a better upper bound of the sum of two negative eigenvalues of a matrix?

If someone can help, I will be grateful.

Given this matrix

$M=\begin{bmatrix} 0 & m-1 & 2 & n-1\\ 1 & m-2& 1 & n-1\\ 2 & m-1 & 0 & 2(n-1)\\ 1 & m-1 & 2 & 2(n-2)\end{bmatrix},$ show that if $\alpha_1,\alpha_2$ are two negative eigenvalues of $M$, then $\alpha_1+\alpha_2<-3$. Alsoo $m,n\ge 4$ are integers.

Using Gershgorin circles theorem, the eigenvalues would lie in the intervals $B(0,m+n)$, $B(m-2,n-1)$, $B(0,2n+m-1)$,$ B(2n-4,m+2)$ where $B(x,y)$ denotes the interval with centre $x$ and radius $y$. But it's not helping to find the required bound?

Does there exist any theorem/trick to find a better upper bound of the sum of two negative eigenvalues of a matrix?

If someone can help, I will be grateful.