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# If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible

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If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible  [#permalink]

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25 May 2020, 01:42
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If $$a^2 + b^2 + c^2 = 1$$, then what is the sum of the minimum possible value and the maximum possible value of $$ab + bc + ca$$?

A. -1/2
B. 0
C. 1/2
D. 1
E. 3/2

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Re: If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible  [#permalink]

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25 May 2020, 06:07
1
Quote:
If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible value and the maximum possible value of ab+bc+ca?

A. -1/2
B. 0
C. 1/2
D. 1
E. 3/2

a^2 + b^2 + c^2 = 1

maximum:

1/2^2+1/2^2+√2/2^2=1
1/4+1/4+1/2=1

1/2*1/2+2(1/2*√2/2)=,
1/4+√2/2=(1+2√2)/4=0.95~

minimum:

1/2^2+1/2^2+(-√2/2)^2=1
1/4+1/4+1/2=1

1/2*1/2+2(1/2)(-√2/2)=,
1/4+(-√2/2)=(1-2√2)/4=-0.45~

-1/2*1/2+(-0.5)*(√2/2)+0.5(√2/2)=√2/2(0
1/4+(-√2/2)=(1-2√2)/4=-0.45~

sum max+min = 0.5

ans (C)
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If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible  [#permalink]

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Updated on: 25 May 2020, 22:40
a^2 + b^2 + c^2 = 1

A few possible sets of (a, b, c) are:
(1/2,1/2,1/√2) --> ab + bc + ac = 1/4 + 1/√2 = 1/4 + 1/1.4 = ~ 1 (max)
(1/√2,0,-1/√2) --> ab + bc + ac = -1/2 (min)

So the sum is 1 - 1/2 = 1/2

Posted from my mobile device

Originally posted by chondro48 on 25 May 2020, 06:22.
Last edited by chondro48 on 25 May 2020, 22:40, edited 1 time in total.
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If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible  [#permalink]

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Updated on: 25 May 2020, 23:02
2
Quote:
If $$a^2+b^2+c^2=1$$, then what is the sum of the minimum possible value and the maximum possible value of ab+bc+ca?

A. -1/2
B. 0
C. 1/2
D. 1
E. 3/2

$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$$
$$(ab+bc+ca)= \frac{(a+b+c)^2 - (a^2 + b^2 + c^2)}{2}$$

as square of a number can not be negative, minimum value is obtained when (a + b + c)^2 is equal to 0.
minimum value = 0 -1 /2 = -1/2

maximum value is obtained when a = b = c = $$\frac{1}{\sqrt{3}}$$
$$(a + b + c)^2 = (\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}})^2 = (\sqrt{3})^2 = 3$$
maximum value = 3 -1/2 = 1

sum of minimum and maximum value = -1/2 + 1 = 1/2
Ans: C

Originally posted by ArunSharma12 on 25 May 2020, 06:33.
Last edited by ArunSharma12 on 25 May 2020, 23:02, edited 1 time in total.
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Re: If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible  [#permalink]

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25 May 2020, 08:42
1
1
square of any number is non-negative

$$(a-b)^2 = a^2+b^2-2ab ≥ 0$$......(1)

$$(b-c)^2 = b^2+c^2-2bc ≥ 0$$........(2)

$$(c-a)^2 = c^2+a^2-2ca ≥ 0$$........(3)

$$2a^2+2b^2+2c^2 - 2ab-2bc -2ca ≥ 0$$

1 ≥ ab+bc+ca

$$(a+b+c)^2 = a^2+b^2+c^2 + 2ab+2bc+2ca ≥ 0$$

$$1+ 2ab+2bc+2ca ≥ 0$$

$$ab+bc +ca ≥ -\frac{1}{2}$$

Sum of maximum and minimum value of ab+bc+ca $$= -\frac{1}{2} +1 = \frac{1}{2}$$

C
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Re: If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible  [#permalink]

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25 May 2020, 22:47
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Re: If a^2 + b^2 + c^2 = 1, then what is the sum of the minimum possible   [#permalink] 25 May 2020, 22:47