Bunuel wrote:

ajit257 wrote:

If a and b are positive, is (a-1 + b-1)-1 less than (a-1b-1)-1?

(1) a = 2b

(2) a + b > 1

Not sure about the ans.

Question: is \((a^{-1}+b^{-1})^{-1}<(a^{-1}*b^{-1})^{-1}\)? --> \((\frac{1}{a}+\frac{1}{b})^{-1}<(\frac{1}{ab})^{-1}\) --> \(\frac{ab}{a+b}<ab\), as \(a\) and \(b\) are positive we can reduce by \(ab\) and finally question becomes: is \(a+b>1\)?

(1) a = 2b --> is \(3b>1\) --> is \(b>\frac{1}{3}\), we don't know that, hence this statement is not sufficient.

(2) a + b > 1, directly gives an answer. Sufficient.

Answer: B.

P.S. ajit257 you should type the question so that it's clear which is an exponent, which is subtraction, and so on.

Hey Bunuel,

once again, this is a little bit fast for me.

I follow your first and third step to reduce the question, but I don't get the second.

I'd explained myself that \((\frac{1}{ab})^{-1}\) = \(1*(\frac{ab}{1})\) so we have ab on the right side.

But I don't follow what you did do reduce the left side. Could you explain in detail?

Thank you!