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

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

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)

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)

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

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

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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