Question

If a, b, c are distinct +ve real numbers and a2 + b2 + c2 = 1, then ab + bc + ca is

greater than 1
equal to 1
less than 1
any real number

Answer 1

Answer:

The correct option is C.

Step-by-step explanation:

In such type of problem if sum of the squares of number is known and you need product of numbers taken two at a time or needed range of the product of numbers taken two at a time. Start with square of the sum of the numbers like

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

2(ab + bc + ca) = (a + b + c)2 – (a2 + b2 + c2)

(ab + bc + ca) = [(a + b + c)2 – 1]/2

Since a2 + b2 + c2 = 1, all a, b and c are less than 1. So, (ab + bc + ca) must be less than 1.

Answer 2

Answer:

.......less than 1

hope it's helps...