proving that $\alpha_1+\alpha_2<-3$ which are the negative roots of $f(x)$

We have \begin{align} f(-3) &= mn + 2m + 5n + 2 > 0, \\ f(-3/2) &= \frac{-20mn + 26m - 4n + 5}{16} < 0,\\ f(0) &= mn - m - n - 7 > 0, \\ f(m) &= -m^3 n+m^3-5 m^2 n+3 m^2-7 m n-m-n-7 < 0, \\ f(2m + 2n) &= 8 m^4+28 m^3 n+32 m^2 n^2+12 m n^3+28 m^3+78 m^2 n+66 m n^2\\ &\quad +16 n^3+26 m^2+51 m n+24 n^2-m-n-7\\ & > 0. \end{align} (Note: Simply letting $m = 4 + s, n = 4 + t$ for $s, t \ge 0$, all the inequalities are obvious.)

Thus, $f(x) = 0$ has four real roots $x_1, x_2, x_3, x_4$ located in the following intervals respectively $$x_1 \in (-3, -3/2), \quad x_2 \in (-3/2, 0), \quad x_3 \in (0, m), \quad x_4 \in (m, 2m+2n).$$

We need to prove that $x_1 + x_2 < -3$ or $x_2 < -3 - x_1$.

Since $-3 - x_1 \in (-3/2, 0)$, it suffices to prove that $f(-3-x_1) > 0$.

Since $f(x_1) = 0$, it suffices to prove that $f(-3-x_1) - f(x_1) > 0$ that is $$(2x_1 + 3)\Big((m+2n)x_1^2 + 3(m+2n)x_1 + m+2n + 3\Big) > 0.$$

It suffices to prove that $$x_1 > - \frac{3}{2} - \sqrt{\frac{5m + 10n - 12}{4m + 8n}}.$$

It suffices to prove that $$f\left(- \frac{3}{2} - \sqrt{\frac{5m + 10n - 12}{4m + 8n}}\right) > 0$$ that is $$\frac{m^3+2 m^2 n+2 m n^2+4 n^3-m^2-10 m n-16 n^2+3 m+6 n+9}{(m+2 n)^2} > 0$$ which is true. (Note: Simply letting $m = 4 + s, n = 4 + t$ for $s, t\ge 0$, this inequality is obvious.)

We are done.

$\phantom{2}$

For @Math_Freak: The Maple code for the last two equations is given by

f := x^4+(-m-2*n+6)*x^3+(m*n-5*m-8*n+10)*x^2+(3*m*n-7*m-8*n)*x+m*n-m-n-7
x := -3/2+Q
f1 := collect(expand(f), Q)
f2 := subs({Q^2 = (5*m+10*n-12)/(4*m+8*n), Q^3 = (5*m+10*n-12)*Q/(4*m+8*n), Q^4 = ((5*m+10*n-12)/(4*m+8*n))^2}, f1)
factor(f2)

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.