If a , b , c are distinct positive real numbers and a2 + b2 + c2a2+b2+c2=1 =1, then ab+bc+caab+bc+ca is

Question

If a , b , c are distinct positive real numbers and ${\mathrm{a2\; +\; b2\; +\; c2}}^{}$ =1, then $ab+bc+ca$ is
1. less than 1
2. equal to 1
3. greater than 1
4. any real no

Answer 1

Correct Answer - Option 1 : less than 1

Calculation:

As we know (a - b)² + (b - c)² + (c - a)² > 0

⇒ a² + b² - 2ab + b² + c² - 2bc + c² + a² - 2ac > 0

⇒ 2(a2 + b2 + c²) -2(ab + bc + ca) > 0

∵ $a^2+b^2+c^2=1$ 

⇒ 2(1) -2(ab + bc + ca) > 0

⇒ 1 - (ab + bc + ca) > 0

⇒ ab + bc + ca < 1

In such type of problem if the sum of the squares of numbers is known and you need a product of numbers taken two at a time or needed range of the product of numbers taken two at a time. Then try to find an algebraic formula where you can find both the term either on LHS or RHS.