TITLE: proving that $\alpha_1+\alpha_2<-3$ which are the negative roots of $f(x)$
QUESTION [11 upvotes]: 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 $m,n$?

REPLY [5 votes]: 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)