\((a+c)(b+d) = 1\)

inequality
Published

February 1, 2025

Find all real numbers \(t\ge 1\) that satisfy the following condition: for any \(a,b \in [-1,t]\) , there always exist \(c,d \in [-1,t]\) such that \((a+c)(b+d) = 1\).

求出所有满足下面要求的不小于1的实数t，对任意\(a,b \in [-1,t]\) 总存在\(c,d \in [-1,t]\) 使得\((a+c)(b+d) = 1\).

Source: 2023年高联A卷一试

We first show that \(\frac{3}{2} \le t \le 2\) is the necessary condition.

Apparently \(t \le 2\). Otherwise \((t+c)(t+d) \ge (t-1)^2 > 1\).

Consider \(a=-1\), \(0<b<1\) and \(c,d \in [-1,t]\) such that \((c-1)(b+d) = 1\).

If \(c-1 < 0\), then \(b+d = \frac{1}{c-1} \le -\frac{1}{2} \implies b \le -\frac{1}{2} -d \le \frac{1}{2}\).

If \(c-1 > 0\), then \(b = \frac{1}{c-1} -d \implies b \ge \frac{1}{t-1} -t\).

Intuitively, the smallest \(t\) should be the value such that \(\frac{1}{t-1} -t = \frac{1}{2}\), i.e. \(t \ge \frac{3}{2}\). It is indeed the case since if \(t < \frac{3}{2}\), then for \(b \in (\frac{1}{2}, \frac{1}{t-1} -t)\), there is no \(c,d \in [-1,t]\) such that \((-1+c)(b+d) = 1\).

We then show that \(\frac{3}{2} \le t \le 2\) is the sufficient condition.

  1. Case 1: \(-1 \le a \le 0\), \(-1 \le b \le 0\).

    2. Just let \((c,d) = (-1 -a, -1-b)\).

  2. Case 2: \(0 \le a \le t\), \(0 \le b \le t\).

    3. Just let \((c,d) = (1 -a,1 -b)\).

  3. Case 3: \(-1 \le a \le 0\), \(0 \le b \le t\)
    • Case 3A: \(-1 \le a \le 0\), \(\frac{1}{2} \le b \le t\)

      • Note that for any \((a,b)\) in this rectangle region, there exists \(c \in [\frac{1}{2} , \frac{3}{2} ]\),\(d \in [2-t,\frac{3}{2}]\) such that \((a+c, b+d) = (\frac{1}{2}, 2)\) because \[(-1, \frac{1}{2}) + (\frac{3}{2}, \frac{3}{2}) = (0, \frac{1}{2}) + (\frac{1}{2}, \frac{3}{2}) = (-1, t) + (\frac{3}{2}, 2 - t) = (0, t) + (\frac{1}{2}, 2 - t) = (\frac{1}{2}, 2)\]

    • Case 3B: \(-1 \le a \le 0\), \(\frac{1}{a-1} + 1 \le b \le \frac{1}{2}\), i.e. \((a,b)\) is in the region bounded by \((x-1)(y-1) = 1, x=0\) and \(y = \frac{1}{2}\).
      • Let \(c=\frac{3}{2}, d = \frac{2}{2a+3} -b\), note that \(d\) is in \([\frac{2}{3}, \frac{3}{2}]\) which is feasible.
    • Case 3C: \(-1 \le a \le 0\), \(0 \le b \le \frac{1}{a-1} + 1\), i.e. \((a,b)\) is in the region bounded by \((x-1)(y-1) = 1, x=-1\) and \(y = 0\).
      • Let \(c=-1, d = \frac{1}{a-1} -b\), note that \(d\) is in \([-1, \frac{1}{2}]\) which is feasible.
  4. Case 4: \(0 \le a \le t\), \(-1 \le b \le 0\)

    5. Note that Case 3 and Case 4 are symmetric with respect to the line \(y=x\), hence this case also works.

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