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Which of the following inequalities is always true for any

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Which of the following inequalities is always true for any [#permalink]

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New post 09 Nov 2012, 11:03
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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|
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Re: Which of the following inequalities is always true for any [#permalink]

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New post 09 Nov 2012, 12:32
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It seems such an easy answer, but I would go with (C). No thinking here bro, you got to know one of the fundamental properties of absolute value, which is subadditivity in a nutshell means that sum of two any elements is something (some number) which is less than or equal to the sum of the each element taken on itself separately, i.e. |a + b| <= |a| + |b|

Please, correct me if I went awry
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Re: Which of the following inequalities is always true for any [#permalink]

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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.
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New post 20 Nov 2012, 10:10
Please refer this brilliant resource (it includes the above mentioned properties and other relevant information)

gmat-math-book-in-downloadable-pdf-format-130609.html
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Re: Which of the following inequalities is always true for any [#permalink]

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New post 06 Dec 2012, 01:17
:wink:

One must memorize and understand this property by heart.
\(|x| + |y| >= |x+y|\)
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New post 07 Jan 2013, 08:21
What is the prove for this properties? Can someone prove this? Pleaseee
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Re: Which of the following inequalities is always true for any [#permalink]

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New post 08 Jan 2013, 03:57
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KevinBrink wrote:
What is the prove for this properties? Can someone prove this? Pleaseee


Square \(|x+y|\leq{|x|+|y|}\) (we can safely do that since both sides are non-negative):

\(x^2+2xy+y^2\leq{x^2+2|xy|+y^2}\) --> \(2xy\leq{2|xy|}\) --> \(xy\leq{|xy|}\) --> always true.

The same way for the second inequality.
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Re: Which of the following inequalities is always true for any [#permalink]

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New post 16 Dec 2013, 12:29
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.
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Re: Which of the following inequalities is always true for any [#permalink]

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Re: Which of the following inequalities is always true for any [#permalink]

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New post 02 Apr 2016, 01:24
Hello from the GMAT Club BumpBot!

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Re: Which of the following inequalities is always true for any   [#permalink] 02 Apr 2016, 01:24
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