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Re: 1/a+1/b+1/c or abc/ab+bc+ac [#permalink]
Expert Reply
Let


\(S=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+a c}{a b c}\)



Then Quantity A is S

Quantity B is \(\frac{a b c}{a b+b c+a c}=\frac{1}{S}\) .

Without any constraints on $a, b, c$ (even if positive), $S$ could be greater than, less than, or equal to 1. For example:


If a=b=c=1 , then S=3 , so A > B.
If a=b=c=3 , then S=1 , so A = B.
If a=b=c=4 , then S=0.75 , so A < B.


Therefore, no fixed comparison exists.

D
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Re: 1/a+1/b+1/c or abc/ab+bc+ac [#permalink]
1
Expert Reply
masudkamrul wrote:
I think the Answer is D

===================================================================
SOLUTION BREAKDOWN
===================================================================

STEP 1: SIMPLIFY QUANTITY A
-------------------------------------------------------------------
Quantity A: 1/a + 1/b + 1/c

Finding the common denominator (abc):
Quantity A = (bc + ac + ab) / abc
= (ab + bc + ac) / abc


STEP 2: COMPARE QUANTITY A AND QUANTITY B
-------------------------------------------------------------------
Quantity A = (ab + bc + ac) / abc
Quantity B = abc / (ab + bc + ac)

Observation: Quantity B is the reciprocal (inverse) of Quantity A.


STEP 3: TEST VALUES
-------------------------------------------------------------------
Since no constraints are given for a, b, and c, we can plug in numbers:

Case 1: Let a = 2, b = 2, c = 2
- Quantity A = 1/2 + 1/2 + 1/2 = 1.5
- Quantity B = 1 / 1.5 = 0.67
- Outcome: Quantity A is greater (1.5 > 0.67)

Case 2: Let a = 6, b = 6, c = 6
- Quantity A = 1/6 + 1/6 + 1/6 = 3/6 = 0.5
- Quantity B = 1 / 0.5 = 2
- Outcome: Quantity B is greater (2 > 0.5)


CONCLUSION
-------------------------------------------------------------------
Because the outcome changes depending on the values selected,
the relationship cannot be determined.

CORRECT ANSWER:
D) The relationship cannot be determined from the information given.
===================================================================


Yes. Updated the OA
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Re: 1/a+1/b+1/c or abc/ab+bc+ac [#permalink]
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