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# If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ?

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Math Expert
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If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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15 Jan 2015, 05:59
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If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ?

(1) |a| + |b| = a + b

(2) a > b

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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17 Jan 2015, 05:36
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Statement (1): From the information given, we can conclude that both, a and b are not negative. The two absolute values must be positive (the question stem tells us they are not 0), so a + b must also be positive. As the absolute values have the same sign, a and be must also have the same sign, otherwise their sum would not be equal to the sum of their absolute values.

Picking a few number pairs, we quickly realize that that information is sufficient. This also makes sense theoretically. Going from $$\frac{1}{a}$$ to $$\frac{1}{(a+b)}$$ will always decrease the term for positive a and b, while going from $$\frac{1}{a}$$ to $$\frac{1}{a}+\frac{1}{b}$$ will always increase the term.

So statement 1 is sufficient.

Statement (2): Here we can easily show that this is not sufficient by picking numbers. Choose e.g. 2 and 2, so we get $$\frac{1}{4}<1$$, which is obviously true. But picking 2 and -1 will give $$1<\frac{1}{2}-1$$, which is obviously wrong.

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Re: If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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25 Jan 2015, 11:38
Lets Consider stmt1:
From stmt1 we can say that both a and b are positive. Then only |a| + |b| = a + b .
now form the question we have to prove that
1/(a+b) < 1/a + 1/b
or we can rephrase that we have to show that
[ab-(a+b)^2]/ab(a+b) < 0

In this case denominator cannot be less than zero as since a and b are positive, ab has to be positive and so do a + b
so the numerator has to be less than zero
so ab-(a+b)^2 < 0
which comes down to a^2 + b ^2 > -ab
which will always be true.

We are able to derive a conclusion using statement 1 .

If we consider 2. Straight away we can say that it's not sufficient.

Hence Option A is the answer.

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If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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07 Feb 2016, 22:36
littlewarthog wrote:
Statement (1): From the information given, we can conclude that both, a and b are not negative. The two absolute values must be positive (the question stem tells us they are not 0), so a + b must also be positive. As the absolute values have the same sign, a and be must also have the same sign, otherwise their sum would not be equal to the sum of their absolute values.

Picking a few number pairs, we quickly realize that that information is sufficient. This also makes sense theoretically. Going from $$\frac{1}{a}$$ to $$\frac{1}{(a+b)}$$ will always decrease the term for positive a and b, while going from $$\frac{1}{a}$$ to $$\frac{1}{a}+\frac{1}{b}$$ will always increase the term.

So statement 1 is sufficient.

Statement (2): Here we can easily show that this is not sufficient by picking numbers. Choose e.g. 2 and 2,so we get $$\frac{1}{4}<1$$, which is obviously true. But picking 2 and -1 will give $$1<\frac{1}{2}-1$$, which is obviously wrong.

1) a and B has to be positive forIaI+IbI=a+b to be correct .
so any posive value will satisfy the question.
sufficient.
2) a>b

consider both to be positive in 1st case and negative in 2nd case ,we get different answers.
a=2 & b=1 .then yes.
a=-2 & b=-3 then no.
Not sufficient.

Ans A.

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Re: If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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07 Feb 2016, 23:23
Bunuel wrote:
Bunuel wrote:
If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ?

(1) |a| + |b| = a + b

(2) a > b

Kudos for a correct solution.

when I rearrange the question stem, I get:
ab<(a+b)^2

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Math Expert
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Re: If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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07 Feb 2016, 23:59
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robu wrote:
Bunuel wrote:
Bunuel wrote:
If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ?

(1) |a| + |b| = a + b

(2) a > b

Kudos for a correct solution.

when I rearrange the question stem, I get:
ab<(a+b)^2

Hi robu,
You cannot cancel out term on both sides by cross multiplication..
you get the two terms on the same side and then simplify..
1/(a+b)< 1/a + 1/b..
1/(a+b)< (a+b)/ab..
1/(a+b)- (a+b)/ab <0..
{ab-(a+b)^2}/ab(a+b)<0...
now you see, you have two situations..
where {ab-(a+b)^2}<0 and ab(a+b)>0...
and where {ab-(a+b)^2}>0 and ab(a+b)<0...

you can furhter simplify, but your observation may not be correct, when you are dealing with inequality and variables..
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Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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25 Feb 2016, 20:17
when I rearrange the question stem, I get:
ab<(a+b)^2

it means we can not cross multiply in inequalities question? please clarify.[/quote]

robu

one can cross multiply if one is sure about signs of both sides.
but in my opinion it is better not to cross multiply....

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If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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13 Sep 2016, 09:46
A.

(1) |a| + |b| = a + b tells us that both values a and b are positive; otherwise, |a| + |b| ≠ a + b if any value is a negative. Therefore, knowing the variables are positive, you can manipulate the equation. 1/(a+b) < (a+b)/ab. Cross multiply and since numerator values are positive, direction of sign is maintained: ab<(a+b)(a+b). Here you can see that the right side will become a^2 +2ab+b^2, definitely greater than ab.

(2) a>b, too many scenarios if a is positive b is negative, or both positive, and one positive one negative.

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Re: If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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03 Dec 2016, 19:28
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what is given?
AB= 0
hence both A and B are non zero but they can be both positive and negative that we don't know

and a and b also do not have the same value, They are distinct numbers.

Look at statement 1 |A|+|B|=A+B ie, the distance of A and B is positive. Hence A and B are positive numbers. Pick any positive number you will get a strict YES or NO answer.
SUFFICIENT

Statement 2: Does not give us any idea about the sign of the variable it just shows a comparision. NS

Experts Do let me know if I am correct. I opt simple ways to save time.

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Re: If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ? [#permalink]

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27 Dec 2017, 16:55
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Re: If ab ≠ 0 and a + b ≠ 0, is 1/(a+b)< 1/a + 1/b ?   [#permalink] 27 Dec 2017, 16:55
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