Let $x$, $y$, and $z$ be nonzero real numbers. Find all possible values of

\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|}

sandwich Dec 9, 2023

#1**0 **

We can consider each of the 8 possible cases for the signs of x,y,z, and x+y+z.

Case 1: x,y,z,x+y+z are all positive. In this case, the fractions ∣x∣x,∣y∣y,∣z∣z,∣x+y+z∣x+y+z all reduce to 1, so the value is 4.

Case 2: x,y,z are all positive and x+y+z is negative. In this case, ∣x∣x=∣y∣y=∣z∣z=∣x+y+z∣x+y+z=−1, so the value is −4.

Case 3: x,y,z are all positive and only one of them is negative. We can say without loss of generality that x is negative. In this case, ∣x∣x=∣y∣y=∣z∣z=1 while ∣x+y+z∣x+y+z=−1, so the value is 1.

Case 4: x,y,z are all negative and x+y+z is positive. In this case, ∣x∣x=∣y∣y=∣z∣z=−1 while ∣x+y+z∣x+y+z=1, so the value is −1.

Case 5: x and y are positive and z is negative. In this case, ∣x∣x=∣y∣y=1 while ∣z∣z=−1 and ∣x+y+z∣x+y+z=x+y+zx+y−z, where without loss of generality we can assume y>z, so x+y+zx+y−z<1. Therefore, the value is between 1 and 2.

Case 6: x and z are positive and y is negative. In this case, ∣x∣x=∣z∣z=1 while ∣y∣y=−1 and ∣x+y+z∣x+y+z=x+y+zx+z−y, where without loss of generality we can assume x>y, so x+y+zx+z−y<1. Therefore, the value is between 1 and 2.

Case 7: y and z are positive and x is negative. In this case, ∣y∣y=∣z∣z=1 while ∣x∣x=−1 and ∣x+y+z∣x+y+z=y+z+xy+z−x, where without loss of generality we can assume y>x, so y+z+xy+z−x<1. Therefore, the value is between −1 and 0.

Case 8: x and y are negative and z is positive. In this case, ∣x∣x=∣y∣y=−1 while ∣z∣z=1 and ∣x+y+z∣x+y+z=z+x+yz−x−y, where without loss of generality we can assume y>x, so z+x+yz−x−y<1. Therefore, the value is between −1 and 0.

BuiIderBoi Dec 9, 2023