If a, b, and c are distinct positive numbers and a^2 + b^2 + c^2 = 1

Question

Competition Mode Question

If a, b, and c are distinct positive numbers and  a^2 + b^2 + c^2 = 1  then  ab + bc + ca  is

A. Less than 1 
B. 1 
C. From 1 to 2, not inclusive
D. 2
E. More than 2

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nick1816 · Accepted Answer

AM≥GM \frac{a^2+b^2}{2} > \sqrt{a^2*b^2} .....(1) As a and b are distinct positive numbers, AM≠GM Similarly, \frac{b^2+c^2}{2} > \sqrt{b^2*c^2} .....(2) and \frac{a^2+c^2}{2} > \sqrt{a^2*c^2} ......(3) Add (1), (2) and (3) ab+bc+ca < a^2+b^2+c^2 ab+bc+ca<1

saumyagupta1602 · Answer

(a+b+c)^2= a^2 + b^2 +c^2 + 2 (ab+bc+ca)
(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)
(a-b)^2+(b-c)^2+(c-a)^2>=0
2(a^2+b^2+c^2-ab-bc-ca)>=0, now put a^2+b^2+c^2=1
2(1-ab-bc-ca)>=0
(1-ab-bc-ca)>=0
1>=ab+bc+ca or ab+bc+ca<=1 thus less than 1 should be the correct answer

freedom128 · Answer

If a, b, and c are distinct positive numbers, then a ≠ b ≠ c and therefore,
(a-b)^2 + (a-c)^2 + (b-c)^2 >0
a^2 + b^2 -2*ab + a^2 + c^2 -2*ac + b^2 + c^2 -2*bc > 0
2*(a^2 + b^2 + c^2) -2*(ab + ac + bc) > 0
(a^2 + b^2 + c^2) - *(ab + ac + bc) > 0
1 - (ab + ac + bc) > 0
- (ab + ac + bc) > -1
(ab + ac + bc) < 1

FINAL ANSWER IS (A) Less than 1

exc4libur · Answer

This a^2 + b^2 + c^2 = 1 means that 0 < a,b,c = proper fractions < 1
Largest sum would be when a,b,c=0.5;
0.5*0.5+0.5*0.5+0.5*0.5=0.25*3=0.75<1;

Ans (A)

lacktutor · Answer

If a, b, and c are distinct positive numbers and  a^2+b^2+c^2=1  then  ab+bc+ca :

Let's say that  a = √0.4 ,  b =√0.5  and  c= √0.1 
-->  a^2+b^2+c^2= (√0.4)^{2} + (√0.5)^{2} + (√0.1)^{2} = 0.4+0.5+0.1 =1

-->  ab+bc+ca= (√0.4)(√0.5)+ (√0.5)(√0.1) +(√0.1)(√0.4)= √0.2 +√0.05 +√0.04 =

= √\frac{1}{5} +√\frac{5}{100} +√\frac{4}{100}

Now, let's take a bigger number from that square roots:
-->  √\frac{1}{5} ≈ √\frac{1}{4}= \frac{1}{2} 
-->  √\frac{5}{100} ≈ √\frac{9}{100}=\frac{3}{10} 
-->  √\frac{4}{100}= \frac{2}{10}  
---------------------------
 \frac{1}{2}+ \frac{3}{10} +\frac{2}{10} = 0.5 +0.3+ 0.2 = 1

As you can see, ab+bc+ca is actually less than 1.

Answer(A)

madgmat2019 · Answer

a^2 + b^2 + c^2 = 1 (a-b)^2+(b-c)^2+(c-a)^2>0 a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca>0 2(a^2+b^2+c^2)-2(ab+bc+ca)>0 2(1)-2(ab+bc+ca)>0 ab+bc+ca<2/2 ab+bc+ca<1 OA:A

GMATinsight · Answer

For maximum value of ab+bc+ca, a = b = c = 1/√3 because  a^2+b^2+c^2=1

i.e.  (ab+bc+ca)_{max} = (1/3)+(1/3)+(1/3) = 1

But since a, b and c are distinct therefore, (ab+bc+ca) must be LESS than 1

Answer: Option A

baraa900 · Answer

but how do we get bigger number from the square roots? i mean how did we estimate √1/5 ≈ √1/4

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If a, b, and c are distinct positive numbers and a^2 + b^2 + c^2 = 1

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If a, b, and c are distinct positive numbers and a^2 + b^2 + c^2 = 1

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