What are the maximum and minimum possible values for |x+y|/(|x|+|y|) +

Question

What are the maximum and minimum possible values for |x+y|/(|x|+|y|) + |z+y|/(|z|+|y|) + |z+x|/(x|+|z|)?
A.3 and 1
B.3 and 0
C.4 and 2
D.4 and 0
E. 3 and 2

chetan2u · Answer

Max value of each of the term is 1 each when all x, y and z are of same sign. 
So Max=1+1+1=3

Now the minimum value of each will be when the numerator will be 0. We cannot have each as 0, then the fractions become undefined.

So two of them are negative and one positive OR two negative and one positive where |x|=|y|=|z|
Now two terms where we have a set of negative and positive will be 0 while the third will be 1, so 0+0+1=1

3 and 1

A

Soumya200011 · Answer

chetan2u 
sir can you please elaborate the highlighted part.

sujoykrdatta · Answer

We know that: |x + y| ≤ |x| + |y| for all x and y
- The 'equality' holds if x and y are of the same sign and the 'less than' holds if x and y are of opposite signs

example: if x = -2, y = -3: |x + y| = 5 = |x| + |y|; whereas, if x = -2, y = 3: |x + y| = 1 < |x| + |y| = 5

Thus, to maximize the above, we assume: x, y, z are all positive: 
Thus: |x+y|/(|x|+|y|) + |z+y|/(|z|+|y|) + |z+x|/(x|+|z|) = 1 + 1 + 1 = 3

To minimize, we would try to make all of them '0' by choosing x and y of same magnitude but opposite signs (say, for example: x=2, y = -2)
In that case: 
The first term: x = 2 and y = -2 => |x+y|/(|x|+|y|) = |0|/(|2| + |-2|) = 0
The second term: we take z = -y = 2 => |z+y|/(|z|+|y|) = |0|/(|2| + |-2|) = 0
However, for the third term: |z+x|/(x|+|z|), we already have x = z = 2 => The value = |4|/(|2| + |2|) = 1

Thus, the minimum value = 0 + 0 + 1 = 1

Answer A

BrushMyQuant · Answer

Theory: |A + B| ≤ |A| + |B| 
  - As |A| + |B| will always add up, irrespective of the signs of A and B, as after coming out of Absolute value they will become non-negative
- But, |A +B| can get reduced if both have oppositive signs

=>  Maximum value  of  \frac{|x+y|}{|x|+|y|} + \frac{|y+z|}{|y|+|z|} + \frac{|z+x|}{|z|+|x|}  will be when all x , y and z have the same sign and numerator and denominator will cancel out to give us

Maximum value  = 1 + 1 + 1 = 3

Ideally  Minimum value  should be zero if we had two different variables in each part of the absolute value. But here we have 3 variables shared across three Absolute values. So, two of them will have the same sign and value and other can have a different sign and same value.

Ex: x = 1, y = 1 and z = -1

=>  \frac{|1+1|}{|1|+|1|} + \frac{|1-1|}{|1|+|-1|} + \frac{|-1+1|}{|-1|+|1|} 
=>  Minimum value  =  \frac{2}{2}  + 0 + 0 = 1

So,  Answer will be A 
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

https://www.youtube.com/watch?v=huFiBk8PaGY

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What are the maximum and minimum possible values for |x+y|/(|x|+|y|) +

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What are the maximum and minimum possible values for |x+y|/(|x|+|y|) +

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