Question

If x , y , z are distinct positive real numbers and x² + y² + x² = 1 then xy + yz + zx is
1. > 1
2. - 1
3. < 1
4. Any real number

Answer 1

Correct Answer - Option 3 : < 1

Calculation:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

⇒ 2(xy + yz + zx) = (x + y + z)² – (x² + y² + z²)

⇒ (xy + yz + zx) = [(x² + y² + z²) – 1]/2 < 1

∴ xy + yz + zx is less than 1.