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Can anyone prove the following inequality ?

 
 
Teng
 
Reply Sun 8 Apr, 2012 09:09 am
prove :2008[(1/1000)+(1/1001)+(1/1002)+ ..............+(1/1009)]<20

I deeply appreciated your answer.


Post By Albert.Teng from Taiwan
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Type: Question • Score: 0 • Views: 1,223 • Replies: 7
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MontereyJack
 
  1  
Reply Sun 8 Apr, 2012 09:25 am
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.
Teng
 
  2  
Reply Sun 8 Apr, 2012 09:56 am
@MontereyJack,
Hi MontereyJack:
I am sorry ,your proof is not correct.
altogether we have 10 terms instead of 9 terms

Albert
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MontereyJack
 
  1  
Reply Sun 8 Apr, 2012 10:06 am
true. my mistake.
0 Replies
 
MontereyJack
 
  1  
Reply Sun 8 Apr, 2012 12:37 pm
Brute force, using a calculator proves it's true. Empirical, not elegant, but there you are.
Teng
 
  1  
Reply Sun 8 Apr, 2012 08:37 pm
@MontereyJack,
please prove it without using any computer or calculator.

Albert
0 Replies
 
MontereyJack
 
  1  
Reply Sun 8 Apr, 2012 09:38 pm
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.
Teng
 
  1  
Reply Sun 8 Apr, 2012 11:22 pm
@MontereyJack,
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


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