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Let $a_{i}=i+\frac{1}{i}$ for $i=1,2, \ldots \ldots ., 20 .$ Put $\mathrm{p}=\frac{1}{20}\left(\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots \ldots+\mathrm{a}_{20}\right)$ and $\mathrm{q}=\frac{1}{20}\left(\frac{1}{\mathrm{a}_{1}}+\frac{1}{\mathrm{a}_{2}}+\ldots \ldots+\frac{1}{\mathrm{a}_{20}}\right) .$ Then
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The correct answer is:
$q \in\left(0, \frac{22-\mathrm{p}}{21}\right)$
$\mathrm{q}>0$, try to the contra prove that $\mathrm{q} < \frac{22-\mathrm{p}}{21}$
$\mathrm{q}+\frac{\mathrm{p}}{21} < \frac{22}{21}$
$\mathrm{q}+\frac{\mathrm{p}}{21}=\frac{1}{20}\left[\sum_{\mathrm{i}=1}^{20} \frac{1}{\mathrm{a}_{\mathrm{i}}}+\frac{1}{21} \sum_{\mathrm{i}=1}^{20} \mathrm{a}_{\mathrm{i}}\right]$
$=\frac{1}{20}\left[\sum_{i=1}^{20}\left(\frac{\mathrm{i}}{\mathrm{i}^{2}+1}\right)+\frac{1}{21} \sum_{\mathrm{i}=1}^{20}\left(\mathrm{i}+\frac{1}{\mathrm{i}}\right)\right]$
$=\frac{1}{2}+\frac{1}{20}\left[\sum_{\mathrm{i}=1}^{20} \frac{\mathrm{i}}{\mathrm{i}^{2}+1}+\sum_{\mathrm{i}=1}^{20}\left(\frac{1}{21 \mathrm{i}}\right)\right]$
$=\frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\sum_{\mathrm{i}=2}^{20} \frac{\mathrm{i}}{\mathrm{i}^{2}+1}+\sum_{\mathrm{i}=1}^{20} \frac{1}{21 \mathrm{i}}\right]$
$\quad < \frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\frac{2}{5} \sum_{\mathrm{i}=2}^{20} 1+\frac{1}{21} \sum_{\mathrm{i}=1}^{20} 1\right]$
$ < \frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\frac{2}{5} \times 19+\frac{1}{21} \times 20\right]$
$ < \frac{1}{2}+\frac{1}{20}(1+8+1)$
$ < \frac{1}{2}+\frac{1}{2}$
$ < \frac{22}{21}$
$\mathrm{q}+\frac{\mathrm{p}}{21} < \frac{22}{21}$
$\mathrm{q}+\frac{\mathrm{p}}{21}=\frac{1}{20}\left[\sum_{\mathrm{i}=1}^{20} \frac{1}{\mathrm{a}_{\mathrm{i}}}+\frac{1}{21} \sum_{\mathrm{i}=1}^{20} \mathrm{a}_{\mathrm{i}}\right]$
$=\frac{1}{20}\left[\sum_{i=1}^{20}\left(\frac{\mathrm{i}}{\mathrm{i}^{2}+1}\right)+\frac{1}{21} \sum_{\mathrm{i}=1}^{20}\left(\mathrm{i}+\frac{1}{\mathrm{i}}\right)\right]$
$=\frac{1}{2}+\frac{1}{20}\left[\sum_{\mathrm{i}=1}^{20} \frac{\mathrm{i}}{\mathrm{i}^{2}+1}+\sum_{\mathrm{i}=1}^{20}\left(\frac{1}{21 \mathrm{i}}\right)\right]$
$=\frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\sum_{\mathrm{i}=2}^{20} \frac{\mathrm{i}}{\mathrm{i}^{2}+1}+\sum_{\mathrm{i}=1}^{20} \frac{1}{21 \mathrm{i}}\right]$
$\quad < \frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\frac{2}{5} \sum_{\mathrm{i}=2}^{20} 1+\frac{1}{21} \sum_{\mathrm{i}=1}^{20} 1\right]$
$ < \frac{1}{2}+\frac{1}{20}\left[\frac{1}{2}+\frac{2}{5} \times 19+\frac{1}{21} \times 20\right]$
$ < \frac{1}{2}+\frac{1}{20}(1+8+1)$
$ < \frac{1}{2}+\frac{1}{2}$
$ < \frac{22}{21}$
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