If ab 0 and a + b 0, is 1/(a + b)

Question

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

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

(2) a > b

littlewarthog · Accepted Answer

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.

Therefore, answer A is correct.

Ferrarif1286 · Answer

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.

tcsanimesh · Answer

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) < 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.

rohit8865 · Answer

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.

robu · Answer

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

in above situation, both stems seem sufficient. Is my approach correct? Please help.

chetan2u · Answer

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..

rohit8865 · Answer

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

in above situation, both stems seem sufficient. Is my approach correct? Please help.

it means we can not cross multiply in inequalities question? please clarify.

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....

AmritaSarkar89 · Answer

VERY SIMPLE AND LUCID ANSWER

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

A is the answer:)

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

BunuelWannabe · Answer

Hi  Bunuel

Could you enlighten us with your wisdom in this question?

I cannot elucidate a reliable strategy to get the answer in less than 2:30 mins.

Thanks!

Kritisood · Answer

@experts could you pls help with this question?

chetan2u · Answer

Modify the question.. \frac{1}{(a+b)}< \frac{1}{a} + \frac{1}{b}...........\frac{1}{(a+b)}< \frac{a+b}{ab}. (1) |a| + |b| = a + b this means a+b is positive and also both a and b are positive, otherwise |a| + |b| will never be equal to a + b. So, we can cross multiply our initial question - Is \frac{1}{(a+b)}< \frac{a+b}{ab}..........ab<(a+b)^2.......ab0 . So, we are looking for whether a^2+b^2+ab>0 . We know a and b are positive, so all the terms a^2,b^2,ab will be positive. Hence answer is YES. Sufficient (2) a > b If a>b>0...Yes as shown above in statement I If a>0>b..say 5>0>-1.... \frac{1}{(a+b)}< \frac{1}{a} + \frac{1}{b}...........\frac{1}{(5-1)}< \frac{5-1}{-5}.........\frac{1}{4}< \frac{4}{-5} ...NO Insufficient A

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

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