Bunuel wrote:

derekgmat wrote:

Which of the following inequalities is always true for any real number 'a' and 'b'?

(A) |a + b| = |a| + |b|

(B) |a + b| > |a| + |b|

(C) |a + b| <= |a| + |b|

(D) |a - b| <= |a| - |b|

(E) |a - b| > |a| - |b|

Property worth remembering:

1. Always true:

|x+y|\leq{|x|+|y|}, note that "=" sign holds for

xy\geq{0} (or simply when

x and

y have the same sign);

2. Always true:

|x-y|\geq{|x|-|y|}, note that "=" sign holds for

xy>{0} (so when

x and

y have the same sign)

and |x|>|y| (simultaneously).

Hope it helps.

Let me see if Ive interpreted you correctly:

|x+y|\leq{|x|+|y|} means that "!x! is always denoted in its positive value, and so is !y!, and thus !x+y!, which can take two values, can never be greater than the addition of the two positive values !x! and !y!"

By the same token:

|x-y|\geq{|x|-|y|} means that "since both !x! and !y! in the right hand side are denoted in their positive value, their subtracted value can never be bigger than !x - y!"

Is that interpretation correct? That is how I solved the question.