Q:

Finding a Numerical Solution to a Linear Combination Inequality

Can I use Mathematica to find a value for $\alpha$ that satisfies the below linear combination inequality? If so, how?
$$
\begin{bmatrix}
-1 \\
-1 \\
-1
\end{bmatrix}
<
\begin{bmatrix}
g_1 \\
\vdots \\
g_n
\end{bmatrix}
+
\alpha
\begin{bmatrix}
m_1 \\
\vdots \\
m_n
\end{bmatrix}
<
\begin{bmatrix}
1 \\
\vdots \\
1
\end{bmatrix}
$$
I do not care as much about finding an algebraic expression as I do about finding a numerical solution for $\alpha$ given a set of values $g_1, \dots, g_n$ and $m_1, \dots, m_n$.
Furthermore, an important point is that I would be happy as long as the value of $\alpha$ is such that the above inequality is almost true. In other words, if an exact solution does not exist (which is rather likely) I would like to find a value for $\alpha$ that minimizes the error in the inequality (Currently, I cannot find a more precise way to express what I mean by the "error" in the inequality.)
Update 1: Perhaps one could think of minimizing the "error" in the above inequality in a "least squares sense". After all, if we omit one of the outer vectors, say $[-1 \ -1 \ -1]$, and replace $<$ by $=$, we would have an overdetermined system of equations and could attempt to find a value for $\alpha$ by the Method of Least Squares. 
For example, the system
$$
\begin{bmatrix}
g_1 \\
\vdots \\
g_n
\end{bmatrix}
+
\alpha
\begin{bmatrix}
m_1 \\
\vdots \\
m_n
\end{bmatrix}
=
\begin{bmatrix}
1 \\
\vdots \\
1
\end{bmatrix}
$$
could be rewritten on the form $A\vec{x} = \vec{b}$ as follows:
$$
\begin{bmatrix}
g_1 & m_1 \\
\vdots & \vdots \\
g_n & m_n
\end{bmatrix}
\begin{bmatrix}
1 \\
\alpha
\end{bmatrix}
=
\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}.
$$
Update 2: Since I want all of the entries in my resulting vector to lie between $-1$ and $1$, I believe that I could find a good value for $\alpha$ by solving the subsequent system of $n$ equations in one unknown:
$$
\begin{bmatrix}
g_1 \\
\vdots \\
g_n
\end{bmatrix}
+
\alpha
\begin{bmatrix}
m_1 \\
\vdots \\
m_n
\end{bmatrix}
=
\begin{bmatrix}
0 \\
\vdots \\
0
\end{bmatrix}
$$
I believe that the problem proposed at the very start of my post can be reduced to finding a solution to the above system using the Method of Least Squares. I believe this because $0$ lies precisely in the middle between $-1$ and $1$, and the least squares solution to the above system finds the value of $\alpha$ that will result in all entries of the linear combination being as close to 0 as possible in the "least squares sense". Intuitively, getting all entires as close to zero as possible (or, in other words, as close to precisely in the middle between $-1$ and $1$ as possible) appears to be a very similar problem as the problem of getting all entires to lie within the interval $[-1, 1]$.
Ultimately, I think that solving the overdetermined system
$$
\begin{bmatrix}
m_1 \\
\vdots \\
m_n
\end{bmatrix}
\begin{bmatrix}
\alpha
\end{bmatrix}
=
\begin{bmatrix}
-g_1 \\
\vdots \\
-g_n
\end{bmatrix}
$$
using the Method of Least Squares should also be a solution to the original problem. Can you find any holes in this approach?
Update 3: My proposed approach seems to work fairly well. A test on vectors with 48000 randomly generated entries between $-1$ and $1$ resulted in $7$ entires (~0.15%) that violate the inequality in the resulting linear combination. If I just add together the vectors without using my approach to prepare them, 12 064 entires (~25%) violate the inequality. 

A:

As an example, you could write this quite literally
eqn = {-1, -1, -1} < {3, 4, 5} + α {5, 6, 7} < {1, 1,  1}
(* {-1, -1, -1} < {3 + 5 α, 4 + 6 α, 5 + 7 α} < {1, 1, 1} *)

Reduce[Thread[eqn], α]
(* -(4/5) < α < -(4/7) *)

If your inequalities are infeasible, you will need to decide how you would define a "best" solution.

