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# If positive real numbers a,b,c are in a.p such that abc=4 Then the min

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Math Expert
Joined: 02 Sep 2009
Posts: 58335
If positive real numbers a,b,c are in a.p such that abc=4 Then the min  [#permalink]

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01 Feb 2019, 00:50
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Difficulty:

55% (hard)

Question Stats:

44% (02:26) correct 56% (02:42) wrong based on 21 sessions

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If positive numbers a, b, c are in arithmetic progression such that abc = 4, then the minimum value of b is

A. 2^(1/3)
B. 2^(2/3)
C. 2^(1/2)
D. 2^(3/2)
E. 2^(1/4)

If positive real numbers a,b,c are in a.p such that abc=4 Then the minimum value of b is

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Joined: 02 Aug 2009
Posts: 7940
Re: If positive real numbers a,b,c are in a.p such that abc=4 Then the min  [#permalink]

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01 Feb 2019, 10:48
Bunuel wrote:
If positive numbers a, b, c are in arithmetic progression such that abc = 4, then the minimum value of b is

A. 2^(1/3)
B. 2^(2/3)
C. 2^(1/2)
D. 2^(3/2)
E. 2^(1/4)

If positive real numbers a,b,c are in a.p such that abc=4 Then the minimum value of b is

Two ways..

(I) relation in arithmetic mean and geometric mean
AM$$\geq$$GM = $$\frac{a+b+c}{3}\geq{abc}^{1/3}$$.. Now as a,b,c are in arithmetic progression, a+c=2b , so a+b+c=3b..
$$\frac{3b}{3}\geq{4}^{1/3}........b\geq{2}^{2/3}$$.

(II) let a-b=c-b=d..
so three numbers can be written as (b-d), b, (b+d)..
thus abc = (b-d)* b* (b+d)=$$b(b^2-d^2)=2^2....b^3-bd^2=2^2$$, so $$b^3>2^2...b>2^{2/3}$$
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If positive real numbers a,b,c are in a.p such that abc=4 Then the min  [#permalink]

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02 Feb 2019, 00:27
1
given
abc are in ap
we can write:
a-b=c-b=d..
so three numbers can be written as (b-d), b, (b+d)
abc = (b-d)*b*(b+d)
2^2=b(b^2-d^2)
2^2= b^3-b*d^2

value of b^3>2^2
or say
b>2^2/3

IMO B
chetan2u wrote:
Bunuel wrote:
If positive numbers a, b, c are in arithmetic progression such that abc = 4, then the minimum value of b is

A. 2^(1/3)
B. 2^(2/3)
C. 2^(1/2)
D. 2^(3/2)
E. 2^(1/4)

If positive real numbers a,b,c are in a.p such that abc=4 Then the minimum value of b is

Two ways..

(I) relation in arithmetic mean and geometric mean
AM$$\geq$$GM = $$\frac{a+b+c}{3}\geq{abc}^{1/3}$$.. Now as a,b,c are in arithmetic progression, a+c=2b , so a+b+c=3b..
$$\frac{3b}{3}\geq{4}^{1/3}........b\geq{2}^{2/3}$$.

(II) let a-b=c-b=d..
so three numbers can be written as (b-d), b, (b+d)..
thus abc = (b-d)* b* (b+d)=$$b(b^2-d^2)=2^2....b^3-bd^2=2^2$$, so $$b^3>2^2...b>2^{2/3}$$

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If positive real numbers a,b,c are in a.p such that abc=4 Then the min   [#permalink] 02 Feb 2019, 00:27
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