A few days ago, I encountered an inequality problem: find the range of values for 2a-b, given that: 1 < a + b < 2 \tag{1} -2 < a - b < -1 \tag{2} The teacher happily added the left and right sides of the two equations to get -1 < 2a < 1. Then, multiplying the second equation by (-1) gives 1 < b - a < 2, which when added to (1) yields 2 < 2b < 4 \Rightarrow 1 < b < 2. Next, multiplying this result by (-1) and adding it to -1 < 2a < 1, the result is clearly -3 < 2a - b < 0. Readers, do you think there is a problem with this approach?
First of all, the teacher’s method is rigorous but not precise, and the result obtained is not incorrect. However, we find that we can never find an example where 2a - b = -2.99. This indicates that the range has been expanded. Why? Let’s first look at a very simple example: \begin{aligned} 1 < a < 2 \\ 1 < a < 3 \end{aligned} If the two equations are added, we get 1 < a < 2.5. Obviously, this result expands the range of the first inequality, so this approach is not advisable.
So, how can we be precise? Returning to the original problem, let a + b = x and a - b = y. Then we have: \begin{aligned} 1 < x < 2 \\ -2 < y < -1 \end{aligned} Solving the system of equations: \begin{aligned} a + b = x \\ a - b = y \end{aligned} We get a = \frac{x+y}{2} and b = \frac{x-y}{2}, which implies 2a - b = \frac{x+3y}{2}.
It is easy to see that the larger (or smaller) x and y are, the larger (or smaller) the expression becomes. Substituting x=2, y=-1 for the maximum value and x=1, y=-2 for the minimum value, we get (2a - b) \in (-5/2, -1/2). Now we can see that the answer -3 < 2a - b < 0 was indeed expanded!
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