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How many inequalities are hidden in $|a\pm b| \le |c \pm d|$? Am I right in thinking as follows:

Case 1. $a\pm b >0$ and $c \pm d >0$. In this case we have $a\pm b \le c \pm d$

Case 2. $a\pm b <0$ and $c \pm d >0$. In this case we have $-(a\pm b) \le c \pm d$

Case 3. $a\pm b >0$ and $c \pm d <0$. In this case we have $a\pm b \le -(c \pm d)$

Case 4. $a\pm b <0$ and $c \pm d <0$. In this case we have $-(a\pm b) \le -( c \pm d)$

User101
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    It seems ambiguous because of the $\pm$ signs. For example, the sign of $a\pm b$ can depend on whether $+$ or $-$ is chosen. – Andreas Tsevas Jul 20 '22 at 00:18
  • Yes, @Andreas is right. Your question is impossible to answer $-$ I might even say it's meaningless. Can you restate it without using $\pm$? – TonyK Jul 20 '22 at 00:25

1 Answers1

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There are $(16)$ inequalities hidden in the formula.

The LHS, inside the absolute value signs, may be evaluated as $(a + b),$ or $(a - b)$. The RHS is similar.

So, that is $2^2$ possibilities.

Further, on the LHS, whichever evaluation inside the absolute value signs is selected, this evaluation will either be negative, or non-negative. These cases must be handled separately, because they effect how the absolute value is evaluated. The RHS is again similar.

So, you have a further $2^2$ possibilities.

This implies that there are a total of $2^2 \times 2^2 = 16$ different cases to be considered.

If I was attacking this problem, I would forgo elegance, break the problem into the $16$ cases, and explore each case individually.

I am assuming that it is the problem composer's intent that any $(a,b,c,d)$ that satisfy any one of the $16$ cases should be included in the solution set.

user2661923
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