by **laiusergiu** Sun Jul 24, 2011 5:21 am

I am never comfortable with solutions that resort to number picking.

I spent some time thinking about this problem and here's what I came up with.

Let's start with the fact that |a|+|b|>= |a+b|. This is always the case, with equality occurring when a>=0 and b>=0.

Our question asks us to determine whether |x-y|>|x|-|y|.

|x-y|>|x|-|y| rearrange the terms

|x-y|+|y|>|x| expand x

|x-y|+|y|>|x-y+y| we are now in the |a|+|b|>= |a+b| scenario because the sum of the expressions within the individual absolute values equals the expression under the collective absolute value. Therefore we conclude that this inequality always holds weakly (i.e. >=), with equality occurring when the expressions within the individual absolute values are non-negative (i.e. x-y>=0 and y>=0). In other words

CASE 1: if x>=y>=0 then the left and right sides are equal.

CASE 2: for all other relations between x, y, and 0, the left side will be strictly greater than the right side

Conceptually, this question amounts to us deciding whether we have enough information to unequivocally determine in which of the 2 possible scenarios we are: 1) equality of the two sides or 2) left side is strictly greater than the right side.

Now, let's go through the answer choices:

(1) y<x

This is not enough for us to know whether x>=y>=0 holds. If we also knew that y>=0 then we'd have enough information to state that we are in CASE 1 (i.e. the two sides are equal). In the absence of that piece of information, however, we cannot determine whether we're in CASE 1 or CASE 2.

(2) xy<0

This tells us that x and y are of opposite signs, which implies that x>=y>=0 does not hold, and that puts us in CASE 2 and allows us to conclude that the left side is unequivocally greater than the right side. To be more specific, this allows us to conclude that:

|x-y|+|y|>|x-y+y|

therefore

|x-y|+|y|>|x|

or if you prefer

|x-y|>|x|-|y|

Cheers,

Sergiu