The quadruplet (P, Q, R, S) of positive integers satisfies the relationship P*S = Q*R with the restriction P > Q > R > S

Prove that (P-S)² ≥ 4*S + 8

(In reply to Solution by FrankM)

"But c>=1, b>=2 and b-c>=1 so that (b-c)^2 + 2(b+c) - 7 >= 0"

-> But c>=1, b>=2 and b-c>=1 with b+c >=3 so that (b-c)^2 + 2(b+c) - 7 >= 0.

Very nice solution, which is concise and compact, yet precise. Also, the methodology inclusive of the said solution  overrides the necessity to apply AM – GM inequality.

Consequently, I confirm having hyperlinked this methodology in my solution. 

Edited on May 20, 2008, 1:43 pm

Posted by K Sengupta on 2008-05-20 13:35:01