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

Take P = S+a,  Q = S+b,  R = S+c  with a>b>c then PS = QR implies S = bc/(a-b-c) so clearly a >= b+c+1, and the inequality holds iff

[ (P-S)^2 - 4S - 8 ] [a - b - c] > 0, i.e. iff

a^2 (a-b-c) - 4bc - 8(a-b-c) > 0. This is increasing in a when a>b+c so we can substitute b+c+1 for a without affecting the validity of the inequality. This gives

(b+c+1)^2 - 4bc - 8 = (b-c)^2 + 2(b+c) -7.

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

Edited on May 11, 2008, 7:04 am

Posted by FrankM on 2008-05-11 07:01:18