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

Question

Competition Mode Question

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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nick1816 · Accepted Answer

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)

Add (1), (2) and (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

exc4libur · Answer

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)

freedom128 · Answer

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

FINAL ANSWER IS (C)

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ArunSharma12 · Answer

(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

sambitspm · Answer

See the attachment.
Answer C

bumpbot · Answer

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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

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