Can anyone prove the following inequality ?

simple answer: (1/1000)>(1/1001)>(1/1002)>...(1/1009), so the sum of the series is<(9/1000). So your series will be less than 2008(9/1000)=18,072/1000=18.072<20.

Hi MontereyJack:
I am sorry ,your proof is not correct.
altogether we have 10 terms instead of 9 terms

Albert

true. my mistake.

Brute force, using a calculator proves it's true. Empirical, not elegant, but there you are.

please prove it without using any computer or calculator.

Albert

nope, I've done it twice, once wrong, once right (you didn't specify a method before I proved it). Now it's your turn.

let x=1000 ,then we have:
(2x+8){[1/x + 1/(x+9)]+ [1/(x+1) + 1/(x+8)]+ [1/(x+2) + 1/(x+7)]+-----------------+[1/(x+4) + 1/(x+5)]}<5(2x+8)(2x+9)/(x^2 +9x)
=5(4x^2+34x+72)/(x^2+9x)<5(4x^2+34x+2x+72-72)/(x^2+9x)
=5(4x^2+36x)/(x^2+9x)=5(4)=20#
so we have it
Albert