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Updated on: 06 Mar 2026, 19:34
Originally posted by MartyMurray on 17 Dec 2025, 08:35.
Last edited by MartyMurray on 06 Mar 2026, 19:34, edited 1 time in total.
Last edited by MartyMurray on 06 Mar 2026, 19:34, edited 1 time in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
34% (02:31) correct
66%
(02:30)
wrong
based on 179
sessions
History
Date
Time
Result
Not Attempted Yet
If \(a^2 + 2b^2 + 2ab < c\), which of the following must be true?
I. \(c > 0\)
II. \(c > ab\)
III. \(|c| > |a + b|\)
(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
Source: Marty Murray Coaching
I. \(c > 0\)
II. \(c > ab\)
III. \(|c| > |a + b|\)
(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
Source: Marty Murray Coaching
17 Dec 2025, 10:28
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If \(a^2 + 2b^2 + 2ab < c\), which of the following must be true?
I. \(c > 0\)
For this statement, it's helpful to recognize the following:
\(a^2 + 2b^2 + 2ab < c\)
\((a + b)^2 + b^2 < c\)
So, \(c >\) the sum of two squares.
the sum of two squares \(≥ 0\)
So, \(c > 0\).
I must be true.
II. \(c > ab\)
\(a^2\) and \(b^2\) are both nonnegative.
So, if \(a^2 + 2b^2 + 2ab < c\), then nonnegative \(+ 2ab < c\).
Thus, \(c > ab\).
II must be true
III. \(|c| > |a + b|\)
From \(a^2 + 2b^2 + 2ab < c\), we know that \((a + b)^2 + b^2 < c\) and thus that \(c > (a + b)^2\).
However, it may not be true that \(|c| > |a + b|\) because \(a\) and \(b\) may be small fractions, in which case \((a + b)^2 < |a + b|\).
So, it could be that \(a^2 + 2b^2 + 2ab < c\) even if \(|c| < |a + b|\).
For example, if \(a = b = \frac{1}{8}\), then \(a^2 + 2b^2 + 2ab = \frac{5}{64}\).
In that case, \(c\) could be \(\frac{6}{64}\).
\(|\frac{6}{64}| < |\frac{1}{8} + \frac{1}{8}|\)
III does not have to be true.
(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
Correct answer: C
I. \(c > 0\)
For this statement, it's helpful to recognize the following:
\(a^2 + 2b^2 + 2ab < c\)
\((a + b)^2 + b^2 < c\)
So, \(c >\) the sum of two squares.
the sum of two squares \(≥ 0\)
So, \(c > 0\).
I must be true.
II. \(c > ab\)
\(a^2\) and \(b^2\) are both nonnegative.
So, if \(a^2 + 2b^2 + 2ab < c\), then nonnegative \(+ 2ab < c\).
Thus, \(c > ab\).
II must be true
III. \(|c| > |a + b|\)
From \(a^2 + 2b^2 + 2ab < c\), we know that \((a + b)^2 + b^2 < c\) and thus that \(c > (a + b)^2\).
However, it may not be true that \(|c| > |a + b|\) because \(a\) and \(b\) may be small fractions, in which case \((a + b)^2 < |a + b|\).
So, it could be that \(a^2 + 2b^2 + 2ab < c\) even if \(|c| < |a + b|\).
For example, if \(a = b = \frac{1}{8}\), then \(a^2 + 2b^2 + 2ab = \frac{5}{64}\).
In that case, \(c\) could be \(\frac{6}{64}\).
\(|\frac{6}{64}| < |\frac{1}{8} + \frac{1}{8}|\)
III does not have to be true.
(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III
Correct answer: C
WiziusCareers1
Joined: 27 Apr 2009
Last visit: 24 Apr 2026
Posts: 178
Given Kudos: 35
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Location: India
Schools: HBS '14 (A)
GMAT 1: 730 Q51 V36
02 Jan 2026, 07:30
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To solve this inequality, we should rewrite the expression a^2 + 2b^2 + 2ab into a more recognizable form by completing the square.
a^2 + 2b^2 + 2ab < c can be written as
a^2 + 2ab + b^2 + + b^2 < c
(a + b)^2 + b^2 < c
Minimum value for any square term is 0.
Hence, c > 0.
So, I must be true.
Also, as a^2 + 2b^2 + 2ab < c
c - ab > a^2 + 2b^2 + ab
We can rewrite a^2 + 2b^2 + ab as (a + (b/2))^2 + (7/4)b^2
So, c - ab > (a + (b/2))^2 + (7/4)b^2
Hence, c - ab > 0
=> c > ab
So, II must be true.
Let a = 0.1 and b = 0.1 and let c = 0.06
Now, our inequality holds true.
but |0.06| < |0.1 + 0.1|
So, III is not must be true
The answer is C.
a^2 + 2b^2 + 2ab < c can be written as
a^2 + 2ab + b^2 + + b^2 < c
(a + b)^2 + b^2 < c
Minimum value for any square term is 0.
Hence, c > 0.
So, I must be true.
Also, as a^2 + 2b^2 + 2ab < c
c - ab > a^2 + 2b^2 + ab
We can rewrite a^2 + 2b^2 + ab as (a + (b/2))^2 + (7/4)b^2
So, c - ab > (a + (b/2))^2 + (7/4)b^2
Hence, c - ab > 0
=> c > ab
So, II must be true.
Let a = 0.1 and b = 0.1 and let c = 0.06
Now, our inequality holds true.
but |0.06| < |0.1 + 0.1|
So, III is not must be true
The answer is C.
