39
$\begingroup$

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

$\endgroup$
  • 4
    $\begingroup$ @Downvoters .. care to explain ? =P $\endgroup$ – r9m Jul 1 '14 at 16:18
  • 2
    $\begingroup$ How did you come up with the term $\dfrac{7\log(2)}{8\log(n)}$? $\endgroup$ – robjohn Jul 1 '14 at 19:33
  • 8
    $\begingroup$ This holds for $n=2$. $\endgroup$ – robjohn Jul 1 '14 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. $\endgroup$ – Shakespeare Sep 4 '14 at 14:24
  • 2
    $\begingroup$ math.org.cn/forum.php?mod=viewthread&tid=28961 $\endgroup$ – math110 Dec 26 '14 at 12:26
10
+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!)

$\endgroup$
  • 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 '14 at 11:49
  • 1
    $\begingroup$ I nderstand, I'm working on it right now :) Thanks for the link. $\endgroup$ – cirpis Dec 31 '14 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 '14 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 '15 at 0:08
  • $\begingroup$ Happy New Year to you too :-) and thanks a lot for the wonderful answer !! :D $\endgroup$ – r9m Jan 1 '15 at 5:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.