An exercise

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Mehedi Hasan Nowshad
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An exercise

Unread post by Mehedi Hasan Nowshad » Tue Nov 01, 2016 7:06 pm

Let $a_1,a_2,......a_n$ are positive real numbers such that $\sum_{i=1}^{n} \dfrac{1}{a_i} = 1$. Prove that,

\[ \sum_{i=1}^{n} \dfrac{a_i^2}{i} > \dfrac{2n}{n+1} \]
"Failure is simply the opportunity to begin again, this time more intelligently."
- Henry Ford

SYED ASHFAQ TASIN
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Re: An exercise

Unread post by SYED ASHFAQ TASIN » Thu Nov 03, 2016 1:00 pm

I have a problem in this,have to see the ans!(╯﹏╰)
EARLY TO BED AND EARLY TO RISE,MAKES A MAN HEALTHY,WELTHY AND WISE.[\color]

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asif e elahi
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Location:Sylhet,Bangladesh

Re: An exercise

Unread post by asif e elahi » Fri Nov 04, 2016 12:23 am

Mehedi Hasan Nowshad wrote:Let $a_1,a_2,......a_n$ are positive real numbers such that $\sum_{i=1}^{n} \dfrac{1}{a_i} = 1$. Prove that,

\[ \sum_{i=1}^{n} \dfrac{a_i^2}{i} > \dfrac{2n}{n+1} \]
Wrong for $n=1$.
Hint
Prove $\exists i$ so that $a_i\geq n$.
Also it can be proved that $L.H.S> n^2$

Mehedi Hasan Nowshad
Posts:12
Joined:Sat Jun 13, 2015 1:46 pm
Location:Halishahar, Chittagong

Re: An exercise

Unread post by Mehedi Hasan Nowshad » Fri Nov 04, 2016 8:23 pm

oops! I forgot about the case for n=1. :3
"Failure is simply the opportunity to begin again, this time more intelligently."
- Henry Ford

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