Given a2 + b2 = 1, c2 + d2 = 1, p2 + q2 = 1, where all numbers are real , then
- ab + cd + pq > and equal to 1
- ab + cd + pq < 3
- ab + cd + pq> and equal to 3
- ab + cd + pq<and equal to 3/2
We know that, square is always non-negative.
∴ (a-b)2 + (c-d)2 + (p-q)2 ≥ 0
⇒ a2 + b2 - 2ab + c2 + d2 - 2cd + p2 + q2 - pq ≥ 0
⇒ 1 + 1 + 1 -2(ab + cd + pq) ≥ 0 [Given, a2 + b2 = 1, c2+ d2 = 1 and p2 + q2 = 1]
⇒ 3 ≥ 2(ab + cd + pq)
⇒ ab + cd + pq ≤ 
Since, < 3, so we get
ab + cd + pq ≤ < 3
⇒ ab + cd + pq < 3