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Inequality

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Tolaso J Kos
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Inequality

#1

Post by Tolaso J Kos » Tue Mar 01, 2016 9:18 pm

Let $a_n$ be a sequence of positive real terms. Prove that:

$$\frac{1}{\frac{1}{a_1}}+ \frac{2}{\frac{1}{a_1}+ \frac{1}{a_2}}+ \cdots+ \frac{n}{\frac{1}{a_1}+ \frac{1}{a_2}+\cdots+ \frac{1}{a_n}}< 2\sum_{i=1}^{n}a_i$$
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Re: Inequality

#2

Post by Riemann » Sun Oct 30, 2016 10:05 am

This exercise was proposed at Kömal N journal on November 1998. I give the official solution that makes use of the weighted AM - HM inequality.

\begin{align*}
\sum_{k=1}^{n}\frac{k}{\frac{1}{a_1}+ \frac{1}{a_2}+\cdots \frac{1}{a_k}} &=\sum_{k=1}^{n}\frac{2}{k+1}\cdot \frac{1+2+\cdots+k}{\frac{1}{a_1}+ \frac{2}{2a_2}+ \cdots + \frac{k}{ka_k}} \\
&\leq \sum_{k=1}^{n}\frac{2}{k+1} \cdot \frac{1\cdot a_1 + 2 \cdot 2a_2 +\cdots + k \cdot k a_k}{1+2+\cdots +k} \\
&=\sum_{k=1}^{n} \frac{4}{k\left ( k+1 \right )^2}\sum_{i=1}^{k}i^2 a_i= \sum_{i=1}^{n}i^2 a_i \sum_{k=i}^{n}\frac{4}{k\left ( k+1 \right )^2}\\
&<\sum_{i=1}^{n}i^2 a_i \sum_{k=i}^{n}\frac{2(2k+1)}{k^2 \left ( k+1 \right )^2} \\
&= \sum_{i=1}^{n}i^2 a_i \sum_{k=i}^{n}\left ( \frac{2}{k^2} - \frac{2}{(k+1)^2} \right )\\
&< \sum_{i=1}^{n}i^2 a_i \left ( \frac{2}{i^2}- \frac{2}{\left ( n+1 \right )^2} \right ) \\ &<\sum_{i=1}^{n}i^2 a_i \frac{2}{i^2} \\
&= 2\sum_{i=1}^{n}a_i
\end{align*}
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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