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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

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