Is ac + ((a^2 - 1)(b^2 - 1))^(1/2)

Question

Is  ab + \sqrt{(a^2 - 1)(b^2 - 1)} \leq 1  ?

(1)  a^2 + c^2 = 1 
 
(2)  b^2 + d^2 = 1

Bunuel · Answer

Is  ab + \sqrt{(a^2 - 1)(b^2 - 1)} \leq 1 ? 
 
At first glance, one might be tempted to rewrite the inequality  ab + \sqrt{(a^2 - 1)(b^2 - 1)} \leq 1  as  \sqrt{(a^2 - 1)(b^2 - 1)} \leq 1 - ab , and then square both sides to eliminate the square root. However, we can only square an inequality if we are sure that both sides are non-negative. In this case, since  1 - ab  can be either non-negative or negative, we cannot directly square both sides.
 
If we could establish that  1 - ab  is non-negative, then after squaring the inequality we would need to determine:
 
 Is  (\sqrt{(a^2 - 1)(b^2 - 1)})^2 \leq (1 - ab)^2 ?
 
Is  (a^2 - 1)(b^2 - 1) \leq 1 - 2ab + (ab)^2 ?
 
Is  (ab)^2 - a^2 - b^2 + 1 \leq 1 - 2ab + (ab)^2 ?
 
Is   - a^2 - b^2 \leq - 2ab ?
 
Is   0 \leq a^2 - 2ab + b^2 ?
 
Is   0 \leq (a - b)^2 ? 
 
Since the square of a number is always greater than or equal to 0, the answer to the above question would be YES, regardless of the actual values of  a  and  b . However, this is only true IF we know that  1 - ab  is non-negative.
 
(1)  a^2 + c^2 = 1 
 
Since the square of a number is greater than or equal to 0,  a^2 \leq 1  (if it were greater than 1,  a^2 + c^2  would be greater than 1). This implies that  -1 \leq a \leq 1 . However, we still don't know the value of  b  to determine whether  1 - ab  is non-negative. Thus, this statement alone is not sufficient.
 
(2)  b^2 + d^2 = 1 
 
Similarly, we can deduce from this equation that  -1 \leq b \leq 1 . However, we still don't know the value of  a  to determine whether  1 - ab  is non-negative. Thus, this statement alone is not sufficient.
 
(1)+(2) Given that  -1 \leq a \leq 1  and  -1 \leq b \leq 1 , we can conclude that their product,  ab , must be less than or equal to 1. As a result,  1 - ab  is non-negative. In this case, we know that the answer to the original question is YES. Therefore, combining both statements (1) and (2) is sufficient to answer the question.

Answer: C

umangshah · Answer

Can you explain from where do you infer 1-ab and how it has to be non-negative.. I can understand upto reducing the equation but after that I am clueless  Bunuel

Thanks

Bunuel · Answer

IF  \sqrt{(a^2 - 1)(b^2 - 1)} \leq 1 - ab  is true, "1 - ab" must be non-negative because the left hand side represents the square root of a number, which is either positive or 0. Hence, for "1 - ab" to be greater than or equal to that, it cannot be negative.

waytowharton · Answer

KarishmaB   Bunuel   GMATNinja

Is the answer to this question incorrect? Shouldn't the answer be D?

Reason : From statement 1, we can infer that -1<=a<=1. Once, we know that we can infer that -1<=b<=1 else the value within square root would be negative that is not possible. Similarly, we can deduce the same from Statement 2 as well.

Let me know what am I missing.

KarishmaB · Answer

From statement 1, a could be 1 too (when c = 0). When that happens, b could be any value, say 100. 
Then the required expression will take the value of 100.

(though, good thinking on your part!)

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Is ac + ((a^2 - 1)(b^2 - 1))^(1/2) <= 1 ?

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