proving that $\alpha_1+\alpha_2<-3$ which are the negative roots of $f(x)$ I am stuck on the following problem:

Given
\begin{align*}
f(x) &= x^4 + (-m - 2n + 6)x^3 + (mn - 5m - 8n + 10)x^2\\
&\quad + (3mn - 7m - 8n)x + mn-m-n-7,
\end{align*}
if $\alpha_1,\alpha_2$ are two negative roots of $f(x)=0$, show that $\alpha_1+\alpha_2< -3$ for $m,n\ge 4$.

I took some values of $m,n$ and found the following:
$m=n=4$, $\alpha_1=-2.476, \alpha_2=-0.621$;
$m=5,n=4$, $\alpha_1=-2.49, \alpha_2=-0.59$
I tried some other values of $m,n$,
In all the above cases, I am getting $\alpha_1+\alpha_2<-3$.
But I am not able to prove it. Can someone please help me out?
NOTE:
I found the following. I obtained $-1<\alpha_1<-k$ and $-3<\alpha_2<-3+k$
where $k$ is some function of $m$ and $n$.
I took $m=4,n=5$ and I got $\alpha_1=-2.492<-3+0.55, \alpha_2=-0.592<-0.55$.
I took some other values of $m,n$ and obtained similar stuff.
So I think my guess is correct.
Is there any way to find $k$ such that $k$ is a function of $m,n$?
 A: Let $x_1 < x_2 < 0$ be two negative solutions and I will obtain a bound:
$$x_2 <-\dfrac{3}{10}(5+\sqrt{10})\approx-2.4487.$$
First, rewrite the polynomial as:
\begin{align}
f(x,m,n) = (x^2+3x+1)mn  
\\-(x^3+5x^2+7x+1)m  
\\-(2x^3+8x^2+8x+1)n 
\\+(x^4+6x^3+10x^2-7) 
\end{align}
As my previous update and @RiverLi's comment said, we can obtain the simple bound:
\begin{equation}
\lambda_1 < x_1 < x_2 < \lambda_2 < 0,
\end{equation}
where $\lambda_{1,2} = \dfrac{-3+\sqrt{5}}{2}$ are the roots of the equation $x^2+3x+1 = 0,$ by simply considering $f(\lambda_i,m,n).$ But this is not strong enough because we want an upper bound on $x_1.$ Therefore, for a small $\varepsilon > 0$ to be chosen later, consider the equation:
$$x^2+3x+1 = -\varepsilon\,\,(1)$$
and call its roots $x_{1,2}^{\varepsilon} = \dfrac{-3\pm\sqrt{5-4\varepsilon}}{2},\,x_1^{\varepsilon}<x_2^{\varepsilon} .$ The nice thing about this is that if $x < 0$ is root of $(1),$ then we have a nice form for:
\begin{align}
f(x,m,n) = -\varepsilon mn
\\+(x\varepsilon+2\varepsilon +1)m
\\+(2x\varepsilon+2\varepsilon+1)n
\\+(\varepsilon^2+\varepsilon-7-3x)
\end{align}
which we will call $(2).$
We want to choose $\varepsilon$ so that $f(x_1^{\varepsilon})$ is negative, which will give $x_1 < x_1^{\varepsilon}.$
One can immediately see that:
$$2x_1^{\varepsilon}\varepsilon+2\varepsilon+1 = 1-\varepsilon -\varepsilon\sqrt{5-4\varepsilon}<0\iff 0.344446\approx\varepsilon_0 < \varepsilon < 1.25$$
where $\varepsilon_0$ is the unique positive solution of $1-\varepsilon -\varepsilon\sqrt{5-4\varepsilon}=0.$ Choosing this $\varepsilon = \varepsilon_0$ will give a slightly better bound, but when I was doing it by hand I found $0.35$ works, which gives the quadratic equation:
$$20x^2+60x+27 = 0\implies x_1^{\varepsilon} = -\dfrac{3}{10}(5+\sqrt{10})\approx-2.4487.$$
For this chosen value, the only thing left to show is that $f(x_1^\varepsilon) < 0.$ We automatically know that the coefficient of $n$ is negative by our choice and the free coefficient is trivially negative as well. For the remaining part:
$$-\varepsilon mn+(x\varepsilon+2\varepsilon +1)m \leq m\left(1+x\varepsilon-2\varepsilon\right)=m\cdot\dfrac{2-7\varepsilon - \varepsilon\sqrt{5-4\varepsilon}}{2}<0,$$
which can be verified by hand for $\varepsilon = 0.35.$
To conclude, we then have:
$$x_2+x_1 < -\dfrac{3}{10}(5+\sqrt{10}) +\dfrac{-3+\sqrt{5}}{2}\approx -2.83.$$
But all of this was done to obtain an explicit upper bound on $x_1$, which we previously do not have. So now by considering $x^2+3x+1 = \varepsilon$ for small, suitable $\varepsilon >0,$ I hope to better the not-so-tight upper bound $x_2:$
$$x_2<\lambda_2 = \dfrac{-3+\sqrt{5}}{2}\approx -0.381,$$
next time I got some time on my hand.
A: Let $\alpha_{i,m,n}$ be the $i^\text{th}$ root of $f(x;m,n)$ in ascending order. It is shown that for $m>4,n>4$
$$
\begin{align}
&\alpha_{1,\infty,\infty}<\alpha_{1,m,n}<\alpha_{1,4,4}\\
&\alpha_{2,4,4}<\alpha_{2,m,n}<\alpha_{2,\infty,\infty}
\end{align}
$$
Numerically,
$$
\begin{align}
&-2.618\ldots<\alpha_{1,m,n}<-2.476\ldots\\
&-0.621\ldots<\alpha_{2,m,n}<-0.3819\ldots
\end{align}
$$
leading to the bound
$$
\alpha_{1,m,n}+\alpha_{2,m,n}<-2.858\ldots
$$
The proof is via observing the change of sign of $f$ at the boundaries of the intervals $(\alpha_{1,\infty,\infty}~,\alpha_{1,4,4})$ and $(\alpha_{2,4,4}~,\alpha_{2,\infty,\infty})$, as tabulated below:




$x$
$sgn(f(x,m,n))$




$\alpha_{1,\infty,\infty}$
$+1$


$\alpha_{1,4,4}$
$-1$


$\alpha_{2,4,4}$
$-1$


$\alpha_{2,\infty,\infty}$
$+1$




The infinity roots are $(-3\pm\sqrt{5})/2$.
