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Is there a way to combine the two following sums into only one sum? Perhaps an upper bound?

$$A=\Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^2} \Big) \Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^{1/2}}\Big)^{2}$$

Thank you.

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  • $\begingroup$ Have you tried Cauchy-Schwarz? $\endgroup$ – Eclipse Sun Nov 2 '17 at 14:14
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For positives $x_i$ by Holder we obtain: $$\sum_{i=1}^n\frac{1}{x_i^2}\left(\sum_{i=1}^n\frac{1}{\sqrt{x_i}}\right)^2\geq\left(\sum_{i=1}^n\frac{1}{x_i}\right)^3.$$ Also, by C-S and Chebyshov we obtain: $$\sum_{i=1}^n\frac{1}{x_i^2}\left(\sum_{i=1}^n\frac{1}{\sqrt{x_i}}\right)^2\leq n\sum_{i=1}^n\frac{1}{x_i^2}\sum_{i=1}^n\frac{1}{x_i}\leq n^2\sum_{i=1}^n\frac{1}{x_i^3}.$$

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$A=\Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^2} \Big) \Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^{1/2}}\Big)^{2}$

Let $x_i=x_1+$${i-1}\over{n-1}$$(x_n-x_1)$ where $i=1,2,....n$

Expansion of the sums for Riemann sums:

${x_n-x_1}\over{n-1}$$\Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^2} \Big)$=$\int_{x_1}^{x_n}$ $1\over{x}^2$ $dx$

${x_n-x_1}\over{n-1}$$\Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^{1/2}}\Big)$=$\int_{x_1}^{x_n}$ $1\over{x}^{1/2}$ $dx$ if ${n}$ goes to ${\infty}$.

Completed the integrals and return to A

$A=\Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^2} \Big) \Big(\sum\limits_{i=1}^n \frac{1}{x_{i}^{1/2}}\Big)^{2}$$\le$${4}{(n-1)^3\over{(x_n-x_1)}^2}$ $(\sqrt{x_n}-\sqrt{x_1})^2\over{{x_n}{x_1}}$

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