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25 Apr 2023, 04:24
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Is \(ab + \sqrt{(a^2 - 1)(b^2 - 1)} \leq 1\) ?
(1) \(a^2 + c^2 = 1\)
(2) \(b^2 + d^2 = 1\)
(1) \(a^2 + c^2 = 1\)
(2) \(b^2 + d^2 = 1\)
25 Apr 2023, 04:24
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Official Solution:
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
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
24 Aug 2023, 08:20
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Bunuel
As a general rule, should we also check whether (a^2 - 1)(b^2 - 1) >= 0 ? If this doesn't hold, then the expression would be undefined.
