Q. 8 . Given that
a2 + b2 + c2 = 1
a2 + b2 + c2 - (ab+bc+ca)
≥ 0 as a, b and c are real numbers.
∴ a2 + b2 + c2 ≥ ab + bc + ca
as a2 + b2 + c2 = 1
so ab + bc + ca <1 [from (i)] …............(ii)
Again ½ (a2 + b2 + c2) +ab + bc + ca
= ½ (a2 + b2 + c2 + 2ab + 2bc + 2ca)
= ½ (a + b + c)2
≥ 0
∴ ab + bc + ca > ½ (a2 + b2 + c2)
⇒ ab + bc + ca > -½ x 1
≥ - ½ .............(iii)
∴ (ii) and (iii) gives
-½ ≤ ab + bc + ca ≤ 1
