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I have this inequality that I don't know how to prove. Possibly the inequality between means might be useful. For $n \in \mathbb{N}$:

$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1} \ge 1$

  • 0
    A rather elegant solutions - which combines summands into $2n+1$ pairs and show that each pair has sum larger than $\frac1{2n+1}$ - can be found in [this answer](https://math.stackexchange.com/questions/2781862/inequality-9th-grade-frac1n1-frac1n2-frac13n11/2781882#2781882).2018-05-17

5 Answers 5

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By C-S we obtain: $\sum_{k=1}^{2n+1}\frac{1}{n+k}=\sum_{k=0}^{n-1}\left(\frac{1}{n+1+k}+\frac{1}{3n+1-k}\right)+\frac{1}{2n+1}\geq$ $\geq\sum_{k=0}^{n-1}\frac{(1+1)^2}{n+1+k+3n+1-k}+\frac{1}{2n+1}=\sum_{k=0}^{n-1}\frac{2}{2n+1}+\frac{1}{2n+1}=1.$ Done!

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Write this as $ \sum_{j=1}^{2n+1} \dfrac{1}{n+j} \ge \int_1^{2n+2} \dfrac{dx}{n+x} = \ln \left(\frac{3n+2}{n+1}\right)$

Now $\dfrac{3n+2}{n+1} \ge e$ for $n > \dfrac{e-2}{3-e} \approx 2.549$. Do the cases $n=1$ and $2$ separately.

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We induct on $n$. For $n=1$, it's trivial. Now assume it holds for $n=k$, then $ \frac{1}{(k+1)+1}+\cdots+\frac{1}{3(k+1)}+\frac{1}{3(k+1)+1}=\frac{1}{k+2}+\cdots+\frac{1}{3k+4} $ $ =(\frac{1}{k+1}+\cdots+\frac{1}{3k+1})-\frac{1}{k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4} $ Since the first expression is at least $1$ by the inductive hypothesis, it suffices to show $ \frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\ge \frac{1}{k+1} $ But by Cauchy-Schwarz we obtain $ (3k+2+3k+3+3k+4)(\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4})\ge9 $ Which is equivalent to what we want after dividing by $9(k+1)$. So the inductive hypothesis always holds.

We can also use user8268's hint and apply Cauchy or AM-HM directly: $ (k+1+\cdots+3k+1)(\frac{1}{k+1}+\cdots+\frac{1}{3k+1})\ge (2k+1)^2 $ So it suffices to show $ (2k+1)^2\ge (k+1+\cdots+3k+1)=(1+\cdots+3k+1)-(1+\cdots+k) $ or $ 4k^2+4k+1\ge \frac{(3k+1)(3k+2)}{2}-\frac{k(k+1)}{2}=\frac{8k^2+8k+2}{2}=4k^2+4k+1 $ but this is clearly true.

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    Yes indeed, you are right. Actually i did this exercise the same way you have, but instead of $\frac{1}{k+1}$ I, somehow, got $\frac{1}{k}$ for which this inequality is not true. Thanks!2012-10-28
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In general, if $f$ is a convex function and $p_j$ are non-negative coefficients such that $\sum p_j = 1$ then

$ \sum_j p_j \, f(x_j) \geq f(\sum_j p_j \, x_j). $

(This is Jensen's inequality.) Take $f(x) = 1/x$, $p_j = 1/(2n + 1)$ and $x_j = n + j$ for $j \in \{1, \dotsc, 2n+1 \}$ and your inequality follows.

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It follows immediately from arithmetic mean harmonic mean inequality. Recall that given a set of positive numbers, $\{a_k\}_{k=1}^{k=m}$, we have that $\dfrac{\displaystyle\sum_{k=1}^{m} a_k}m \geq \dfrac{m}{\displaystyle \sum_{k=1}^m \dfrac1{a_k}}$ $\dfrac{\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k}}{2n+1} \geq \dfrac{2n+1}{\displaystyle \sum_{k=1}^{2n+1} (n+k)}$ $\displaystyle \sum_{k=1}^{2n+1} (n+k) = n(2n+1) + \dfrac{(2n+1)(2n+2)}2 = (2n+1)^2$ Hence, $\dfrac{\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k}}{2n+1} \geq \dfrac{2n+1}{(2n+1)^2}$ Hence, $\displaystyle \sum_{k=1}^{2n+1} \dfrac1{n+k} \geq 1$