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Note: I have modified the initial question as follows:

Let $w_1, w_2, \ldots, w_d$ be positive weights, and $x_1, x_2, \ldots, x_d$ be positive variables. Now, let us consider the following harmonic sum $H$ of the $d$ variables w.r.t. the weights:

$H = \frac{1}{\frac{w_1}{x_1} + \frac{w_2}{x_2} + \cdots + \frac{w_d}{x_d}}$.

Then, I am wondering if $H$ can be approximately expressed with some arithmetic formalization as follows:

$H \geq \sum_{i=1}^{\infty} (\tau_i(w_1, \ldots, w_d)\cdot \theta_i(x_1, \ldots, x_d)),$

where each $\tau_i$ (resp. $\theta_i$) is any function independent from the variables $x_1, x_2, \ldots, x_d$ (resp. the weights $w_1, w_2, \ldots, w_d$). The equality would hold when $w_k = x_k$ ($1\leq k \leq d$).

I could not handle this problem in my knowledge. I should be pleased to have any comments or suggestions (on the mathematical field related to this problem).

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  • $\begingroup$ What's wrong with $H\ge \frac 1{\sqrt{ \sum_i w_i^2}} \frac 1{\sqrt {\sum_i x_i^2}}$? $\endgroup$
    – fedja
    Commented Jul 10, 2020 at 2:58
  • $\begingroup$ Thank you very much. That's very helpful for me. And now, I notice that I should have written the equality condition as $w_k = \frac{1}{x_k}$, not $w_k = x_k$. I am sorry for that. My concern is that though the Schwarz inequality holds, when we consider $w_k = \frac{1}{x_k}$ as the expected equality condition, $H$ does not reach the right term; $\frac{1}{\sqrt{\Sigma_i w_i^2}}\frac{1}{\sqrt{\Sigma_i x_i^2}}$. $\endgroup$
    – Yoshitaka
    Commented Jul 10, 2020 at 4:18
  • $\begingroup$ It should be better to modify my question as follows. Suppose that $H$ is written as $\frac{1}{\frac{w_1}{x_1} + \cdots + \frac{w_d}{x_d}}$. Then, is it possible to express $H$ in an arithmetic form like $H \geq \Sigma_i (\tau_i(w_1, \ldots, w_d)\cdot \theta_i(x_1, \ldots, x_d))$ with equality $w_k = x_k (1\leq k \leq d)$, where $\tau_i$ (resp. $\theta_i$) is any function independent from the variables $x_1, \ldots, x_d$ (resp. the weights $w_1,\ldots, w_d$)? $\endgroup$
    – Yoshitaka
    Commented Jul 10, 2020 at 4:41
  • $\begingroup$ I am sorry for the inconvenience to change the initial query (whose solution must be collect with yours). I was happy to have your helpful comment on the problem which troubles me. $\endgroup$
    – Yoshitaka
    Commented Jul 10, 2020 at 4:41

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