43
$\begingroup$

How to show that for $a_1,a_2,\cdots,a_n >0$ real numbers and for $n \ge 3$:

$$\sum_{k=1}^{n}\dfrac{k}{a_{1}+a_{2}+\cdots+a_{k}}\le\left(2-\dfrac{7\ln{2}}{8\ln{n}}\right)\sum_{k=1}^{n}\dfrac{1}{a_{k}}$$

This version seems stronger than the inequality mentioned here.

Addition: A sister problem: For $a_1,a_2,\cdots,a_n >0$ real numbers and for $n \ge 2$, we have the version:

$$\displaystyle \dfrac{1}{1+a_{1}}+\dfrac{1}{1+a_{1}+a_{2}}+\cdots+\dfrac{1}{1+a_{1}+a_{2}+\cdots+a_{n}} \le \sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}$$

The stronger vesrion claims that: For each $n$, $c_n = \left(1-\dfrac{\ln{n}}{2n}\right)$ we have: $$\displaystyle \dfrac{1}{1+a_{1}}+\dfrac{1}{1+a_{1}+a_{2}}+\cdots+\dfrac{1}{1+a_{1}+a_{2}+\cdots+a_{n}} \le c_n\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}$$

$\endgroup$
11
  • 4
    $\begingroup$ @Downvoters .. care to explain ? =P $\endgroup$
    – r9m
    Jul 1, 2014 at 16:18
  • 3
    $\begingroup$ How did you come up with the term $\dfrac{7\log(2)}{8\log(n)}$? $\endgroup$
    – robjohn
    Jul 1, 2014 at 19:33
  • 8
    $\begingroup$ This holds for $n=2$. $\endgroup$
    – robjohn
    Jul 1, 2014 at 21:21
  • 2
    $\begingroup$ @r9m please, where on earth did that constant surface from? It could be seriously helpful depending how you found it. $\endgroup$
    – ShakesBeer
    Sep 4, 2014 at 14:24
  • 2
    $\begingroup$ @coolydudey60 I didn't find the constant, its from here, unfortunately without any answer .. however W|A plot tells us $H_n \le \dfrac{16n\log n}{7(n+1)\log 2}$ for $n \ge 2$. $\endgroup$
    – r9m
    Sep 4, 2014 at 14:36

1 Answer 1

12
+100
$\begingroup$

Haven't had any progress with the original problem but I have done significant progress on the "sister" problem:

Let $A_k=\sum\limits_{i=1}^k a_i $, thus the problem can be re-written as $$\sum\limits_{k=1}^n \frac{1}{1+A_k} \leq \sqrt{\sum\limits_{k=1}^n \frac{1}{a_k}}$$ or, since both sides are positive, $$\left( \sum\limits_{k=1}^n \frac{1}{1+A_k} \right)^2 \leq {\sum\limits_{k=1}^n \frac{1}{a_k}}$$ Now this is very similiar to the Cauchy-Schwarz inequality, that is, for any positive integers $x_1,x_2...x_n$ and $y_1,y_2...y_n$, the following inequality holds: $$\left( \sum\limits_{k=1}^n x_ky_k \right)^2 \leq \left( \sum\limits_{k=1}^n x_k^2 \right)\left( \sum\limits_{k=1}^n y_k^2 \right)$$ So, assuming $x_k=\frac{1}{\sqrt{a_k}}$, we gain that $y_k=\frac{\sqrt{a_k}}{1+A_k}$, now if $ \sum\limits_{k=1}^n y_k^2 =\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2}\leq1$ then the original inequality holds. A very nice proof of this due to r9m himself can be found here.

In case of the stronger version $$\sum\limits_{k=1}^n \frac{1}{1+A_k} \leq c_n\sqrt{\sum\limits_{k=1}^n \frac{1}{a_k}}$$ Using the Cauchy-Schwarz inequality in a similiar fashion we gain that in order for the original inequality to hold, $$\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2}\leq c_n$$ must also hold. To prove this let us look at the maximal values of $\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2}$. To obtain the maxima we must solve a system of $n$ partial derivatives of this sum equal to zero, or notationally:

\begin{cases} &\frac{\partial}{\partial a_1}\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2}=0 \\ &\frac{\partial}{\partial a_2}\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2}=0\\ &.\\ &.\\ &\frac{\partial}{\partial a_n}\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2}=0 \end{cases}

Note that if we define $S(i)=\sum\limits_{k=i}^n \frac{a_k}{(1+A_k)^2}$, then, since no terms of our sum with index less than $i$ contain $a_i$, we may rewrite the system as: \begin{cases} &\frac{\partial}{\partial a_1}S(1)=0 \\ &\frac{\partial}{\partial a_2}S(2)=0\\ &.\\ &.\\ &\frac{\partial}{\partial a_n}S(n)=0 \end{cases}

Note that $S(n)= \frac{a_n}{(1+A_{n-1}+a_n)^2}$, thus $$\frac{\partial}{\partial a_n}S(n)=\frac{1+A_{n-1}-a_n}{(1+A_{n-1}+a_n)^3}$$ Note that it is zero if $a_n=1+A_{n-1}$ Now let us show that in order for maxima to be gained $a_{i}=b_{n-i}(1+A_{i-1})$ for all natural $i<n$ with some sequence $b_i$. So we just gained that $a_{n}=b_0(1+A_{n-2})$ with $b_0=1$. The maximal value of $S(n)$ is thus $$\max(S(n))=\max(\frac{a_n}{(1+A_{n-1}+a_n)^2})=\frac{b_0(1+A_{n-1})}{(1+A_{n-1}+b_0(1+A_{n-1}))^2}=\frac{b_0}{(1+b_0)^2(1+A_{n-1}))}$$

We may now simplify $S(n-1)$, since we know that $a_n=b_0(1+A_{n-1})$, $$S(n-1)=\frac{a_{n-1}}{(1+A_{n-2}+a_{n-1})^2}+S(n)=\frac{a_{n-1}}{(1+A_{n-2}+a_{n-1})^2}+\frac{a_n}{(1+A_{n-1}+a_n)^2}=\frac{a_{n-1}}{(1+A_{n-2}+a_{n-1})^2}+\frac{b_0}{(b_0+1)^2(1+A_{n-2}+a_{n-1})}=\frac{((b_0+1)^2+b_0)a_{n-1}+b_0(1+A_{n-2})}{(b_0+1)^2(1+A_{n-2}+a_{n-1})^2}$$

So the partial derivative: $$\frac{\partial}{\partial a_{n-1}}S(n-1)=\frac{\partial}{\partial a_{n-1}}\frac{((b_0+1)^2+b_0)a_{n-1}+b_0(1+A_{n-2})}{(b_0+1)^2(1+A_{n-2}+a_{n-1})^2}=\frac{((b_0+1)^2-b_0)(1+A_{n-2})-((b_0+1)^2+b_0)a_{n-1}}{(1+b_0)^2(1+A_{n-2}+a_{n-1})^3}$$

Thus for maxima to be atained, $a_{n-1}=b_1(1+A_{n-2})$, and $b_1=\frac{(b_0+1)^2-b_0}{(b_0+1)^2+b_0}$ (particularly, $b_1=\frac 3 5$),

The maximal value of $S(n-1)$ is then $$\max(S(n-1))=\max(\frac{((b_0+1)^2+b_0)a_{n-1}+b_0(1+A_{n-2})}{(1+b_0)^2(1+A_{n-2}+a_{n-1})^2})=\frac{((b_0+1)^2-b_0)(1+A_{n-2})+b_0(1+A_{n-2})}{(b_0+1)^2(1+A_{n-2}+b_1(1+A_{n-2}))^2}=\frac{(b_0+1)^2(1+A_{n-2})}{(b_0+1)^2(b_1+1)^2(1+A_{n-2})^2}=\frac{1}{(b_1+1)^2(1+A_{n-2})}$$

$S(n-2)$ can then be calculated as $$S(n-2)=\frac{a_{n-2}}{(1+A_{n-3}+a_{n-2})^2}+S(n-1)=\frac{a_{n-2}}{(1+A_{n-3}+a_{n-2})^2}+\frac{1}{(b_1+1)^2(1+A_{n-3}+a_{n-2})}=\frac{1+A_{n-3}+(1+(b_1+1)^2)a_{n-2}}{(b_1+1)^2(1+A_{n-3}+a_{n-2})^2}$$

The partial derivative is thus: $$\frac{\partial}{\partial a_{n-2}}S(n-2)=\frac{\partial}{\partial a_{n-2}}\frac{1+A_{n-3}+(1+(b_1+1)^2)a_{n-2}}{(b_1+1)^2(1+A_{n-3}+a_{n-2})^2}=\frac{b_1(b_1+1)(1+A_{n-3})-(1+(b_1+1)^2)a_{n-2}}{(b_1+1)^2(1+A_{n-3}+a_{n-2})^3}$$

So, once again to attain maxima $a_{n-2}=b_2(1+A_{n-3})$ and $b_2=\frac{(b_1+1)^2-1}{(b_1+1)^2+1}$. And the maximum of $S(n-2)$ is $$\max(S(n-2))=\max(\frac{1+A_{n-3}+(1+(b_1+1)^2)a_{n-2}}{(b_1+1)^2(1+A_{n-3}+a_{n-2})^2})=\frac{((b_1+1)^2-1)(1+A_{n-3})+1+A_{n-3}}{(b_1+1)^2(1+A_{n-3}+b_2(1+A_{n-2}))^2}=\frac{(b_1+1)^2(1+A_{n-3})}{(b_1+1)^2(b_2+1)^2(1+A_{n-3})^2}=\frac{1}{(b_2+1)^2(1+A_{n-3})}$$

Notice that the expression for the maximum value of $S(n-2)$ is almost identical to the one with $S(n-1)$, only all the indices are shifted by one, thus all the following operations that will be done on higher indices will be analogous, thus we can say that in general $$\max(S(n-i))=\frac{1}{(1+b_i)^2(1+A_{n-i-1})}$$ $$b_i=\frac{(b_{i-1}+1)^2-1}{(b_{i-1}+1)^2+1}$$ Now notice that $$\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2}=S(1)=S(n-n+1)$$ So $$\max(\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2})=\max(S(n-n+1))=\frac{1}{(1+b_{n-1})^2(1+A_0)}$$ $A_0=0$ since it is the sum of no variables, thus the maxima of $\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2}$ can be expressed as $$\max(\sum\limits_{k=1}^n \frac{a_k}{(1+A_k)^2})=\frac{1}{(1+b_{n-1})^2}$$ Now, let $m(n)=\frac{1}{(1+b_{n-1})^2}$, it can be shown that $m(n)$ follows the recurrence relation $$m(n+1)=\frac{(m(n)+1)^2}{4}$$ So, if $c_n=1-\frac{ln(n)}{2n}$ follows this property, we are done: $$1-\frac{ln(n+1)}{2(n+1)}=\frac{(2-\frac{ln(n)}{2n})^2}{4}$$ $$4-\frac{2ln(n+1)}{n+1}=(2-\frac{ln(n)}{2n})^2$$ $$4-\frac{2ln(n+1)}{n+1}=4-\frac{2ln(n)}{n}+\frac{ln(n)^2}{4n^2}$$ $$\frac{2ln(n+1)}{n+1}=\frac{2ln(n)}{n}-\frac{ln(n)^2}{4n^2}$$ Now this almost holds, since $\frac{2ln(n+1)}{n+1} \sim \frac{2ln(n)}{n}$ and $\frac{ln(n)^2}{4n^2}$ is a decreasing function in the interval $(e;+\infty)$, and its maxima in $(1;+\infty)$ is about $0.034$, in other words very very small, so $\frac{2ln(n+1)}{n+1}$ comes very close to $\frac{2ln(n)}{n}-\frac{ln(n)^2}{4n^2}$, which explains why $1-\frac{ln(n)}{2n}$ was such a good bound, additionally, it is slightly bigger than the first few values of $m(n)$ and I think that it keeps on being just a little bit larger than $m(n)$ as $n$ approaches infinity

Now to get the super sharp boundry, we should find a function satisfying $m(n+1)=\frac{(m(n)+1)^2}{4}$ (On a sidenote: Happy new year!)

$\endgroup$
5
  • 1
    $\begingroup$ The upper bound $1$ can be shown by same method as here, but my main problem was establishing $c_n = \left(1-\frac{\log n}{2n}\right)$ :-) $\endgroup$
    – r9m
    Dec 31, 2014 at 11:49
  • 1
    $\begingroup$ I nderstand, I'm working on it right now :) Thanks for the link. $\endgroup$
    – cirpis
    Dec 31, 2014 at 11:52
  • $\begingroup$ Also I'm not sure if the improved bound is stronger than Cauchy-Schwarz or not .. its just a possibility. I wasn't able to find a counter example though ! ^_^ $\endgroup$
    – r9m
    Dec 31, 2014 at 11:52
  • 1
    $\begingroup$ Solved the "stronger" version. Sorry if this is too messy, but I think this shows the general idea clearly enough. Haven't had any progress with the other inequality though, I will try to use a similiar argument as I did here in a bit. Cheers! $\endgroup$
    – cirpis
    Jan 1, 2015 at 0:08
  • $\begingroup$ Happy New Year to you too :-) and thanks a lot for the wonderful answer !! :D $\endgroup$
    – r9m
    Jan 1, 2015 at 5:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .