If a and b are positive, is (a^-1 + b^-1)^-1 less than (a^-1 * b^-1)^-1 ?
1) a = 2b
2) a + b > 1
Exponents
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- cubicle_bound_misfit
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Simplified,
the q asks for
is ab/(a+b) < ab ? where a, b >0
as (a+b)>0
ineqality becomes is ab < ab(a+b) ? where a,b >0
from 1.
a=2b
q is , IS 2b.b < 2b.b.3b ?
or 1< 3b
if B is a positive fraction >= 1/4 it is not true.
hence 1 not suff.
from stmt 2.
if (a+b) > 1
ab(a+b) > ab for any value of a &b hence Sufficient.
ANSWER : B
regards,
the q asks for
is ab/(a+b) < ab ? where a, b >0
as (a+b)>0
ineqality becomes is ab < ab(a+b) ? where a,b >0
from 1.
a=2b
q is , IS 2b.b < 2b.b.3b ?
or 1< 3b
if B is a positive fraction >= 1/4 it is not true.
hence 1 not suff.
from stmt 2.
if (a+b) > 1
ab(a+b) > ab for any value of a &b hence Sufficient.
ANSWER : B
regards,
Cubicle Bound Misfit
- amitdgr
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cubicle_bound_misfit wrote:
if (a+b) > 1
ab(a+b) > ab for any value of a &b hence Sufficient.
Thanks for reply ... But can you explain this step ... how did you deduce that if (a+b) > 1 , ab(a+b) > ab for all a,b ?
Thanks
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- amitdgr
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The solution just struck me
from the question we get to know that a and b are positive
so ab/(a+b) < ab can be written as ab(a+b) > ab and since ab is positive, we can divide both sides by ab ... that gives us the simplified question
is (a+b) > 1 ?
statement 1 says a = 2b ... this does not help me in telling whether a+b>1
incidentally statement 2 gives us the info directly that (a+b) > 1
so 2 is sufficient
from the question we get to know that a and b are positive
so ab/(a+b) < ab can be written as ab(a+b) > ab and since ab is positive, we can divide both sides by ab ... that gives us the simplified question
is (a+b) > 1 ?
statement 1 says a = 2b ... this does not help me in telling whether a+b>1
incidentally statement 2 gives us the info directly that (a+b) > 1
so 2 is sufficient
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- cubicle_bound_misfit
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- amitdgr
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we know x^-1 = 1/xamitdgr wrote:If a and b are positive, is (a^-1 + b^-1)^-1 less than (a^-1 * b^-1)^-1 ?
1) a = 2b
2) a + b > 1
(a^-1 + b^-1)^-1 can be written as (1/a + 1/b)^-1
further simplified , [(a+b)/ab]^-1 => [ ab/(a+b) ]
hope that helps
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