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I have been struggling for about 5 days to solve this problem, but, so far, I don't even have a starting point. Could you give me a hint, or tell me how you would approach it?

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    $\begingroup$ What have you tried? It looks like the Cauchy-Schwarz inequality must be involved. $\endgroup$ Oct 28, 2015 at 5:00
  • $\begingroup$ Source of this problem? $\endgroup$
    – user5402
    Oct 31, 2015 at 19:19
  • $\begingroup$ I don't know the exact source, one friend told me about it. $\endgroup$
    – cristid9
    Oct 31, 2015 at 21:19

1 Answer 1

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Denote: $S_a=\sum\limits_{k=1}^{n}t_k^a$, for $a \in \{-1,1,2\}$

Given: $S_2S_{-1}=2S_1,S_1S_{-1}=\frac{3n^2}{2}$ and $H(t_1,t_2,...t_n)=\frac{n}{S_{-1}}$

We need to show that: $$\sum^{n}_{k=1} \frac{t_k^3}{S_2-t_k^2} \geq \frac{n^2}{2S_{-1}}$$ We have: $$\sum\limits_{k=1}^{n} \frac{t_k^3}{S_2-t_k^2}=\sum\limits_{k=1}^{n} \frac{t_k^2}{\frac{S_2}{t_k}-t_k} \geq \frac{\bigg( \sum\limits_{k=1}^{n} t_k\bigg)^2}{\sum\limits_{k=1}^{n}\frac{S_2}{t_k}-\sum\limits_{k=1}^{n}t_k}$$ $$=\frac{S_1^2}{S_2S_{-1}-S_1} =\frac{S_1^2}{2S_1-S_1}=S_1>\frac{S_1}{3}=\frac{n^2}{2S_{-1}}$$

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  • $\begingroup$ What does $H(...)$ mean? $\endgroup$
    – cristid9
    Nov 1, 2015 at 16:51
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    $\begingroup$ In your problem, $H(...)$ denotes the harmonic average of $t_1,t_2,...t_n$ $\endgroup$
    – SiXUlm
    Nov 1, 2015 at 19:54

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