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Question

Ifa,bandc are three distinct positive real numbers which are in H.P, then 3a+2b2a-b+3c+2b2c-bis


A

greater than or equal to10

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B

less than or equal to10

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C

only equal to10

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D

none of these

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Solution

The correct option is A

greater than or equal to10


Step 1: Apply the H.P relationship ona,bandc.

We have given, a,b&c are in H.P

and we know that

b=2aca+cb(a+c)=2ac

Step 2: Simplifying the given expression.

3a+2b2a-b+3c+2b2c-b=(3a+2b)(2c-b)+(3c+2b)(2a-b)(2a-b)(2c-b)=6ac-3ab+4bc-2b2+6ac-3bc+4ab-2b24ac-2ab-2bc+b2=12ac+b(a+c)-4b24ac-2b(a+c)+b2=12ac+2ac-4b24ac-2(2ac)+b2=14ac-4b2b2=14acb2-4

Step 3: Replace2acb=a+c.

7(a+c)b-4ā‡’7(a+c)Ɨ22Ɨb-4āˆµa,bandcareinH.Pāˆ“2b=1a+1cā‡’7(a+c)2(1a+1c)-4ā‡’72(1+ac+ca+1)-4ā‡’72(2+ac+ca)-4ā‡’72(ac+ca)+3

Step 4: Apply A.M and G.M inequality onacandca.

on applying A.M and G.M relation on acandca

ac+caā‰„2acƗcaac+caā‰„2

now,

ā‡’72(ac+ca)ā‰„72Ɨ2ā‡’3+72(ac+ca)ā‰„7+3ā‡’3+72(ac+ca)ā‰„10

Hence, the option(A) is correct.


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