Vrftsjtryk

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

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

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:
$$displaystyle dfrac{1}{1+a_{1}}+dfrac{1}{1+a_{1}+a_{2}}+cdots+dfrac{1}{1+a_{1}+a_{2}+cdots+a_{n}} le c_nsqrt{dfrac{1}{a_{1}}+dfrac{1}{a_{2}}+cdots+dfrac{1}{a_{n}}}$$

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

Let $A_k=sumlimits_{i=1}^k a_i $, thus the problem can be re-written as
$$sumlimits_{k=1}^n frac{1}{1+A_k} leq sqrt{sumlimits_{k=1}^n frac{1}{a_k}}$$
or, since both sides are positive,
$$left( sumlimits_{k=1}^n frac{1}{1+A_k} right)^2 leq {sumlimits_{k=1}^n frac{1}{a_k}}$$
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:
$$left( sumlimits_{k=1}^n x_ky_k right)^2 leq left( sumlimits_{k=1}^n x_k^2 right)left( sumlimits_{k=1}^n y_k^2 right)$$
So, assuming $x_k=frac{1}{sqrt{a_k}}$, we gain that $y_k=frac{sqrt{a_k}}{1+A_k}$,
now if $ sumlimits_{k=1}^n y_k^2 =sumlimits_{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
$$sumlimits_{k=1}^n frac{1}{1+A_k} leq c_nsqrt{sumlimits_{k=1}^n frac{1}{a_k}}$$
Using the Cauchy-Schwarz inequality in a similiar fashion we gain that in order for the original inequality to hold,
$$sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2}leq c_n$$ must also hold.
To prove this let us look at the maximal values of $sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2}$. 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}sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2}=0 \
&frac{partial}{partial a_2}sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2}=0\
&.\
&.\
&frac{partial}{partial a_n}sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2}=0
end{cases}

Note that if we define $S(i)=sumlimits_{k=i}^n frac{a_k}{(1+A_k)^2}$, 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
$$sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2}=S(1)=S(n-n+1)$$
So
$$max(sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2})=max(S(n-n+1))=frac{1}{(1+b_{n-1})^2(1+A_0)}$$
$A_0=0$ since it is the sum of no variables, thus the maxima of $sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2}$ can be expressed as
$$max(sumlimits_{k=1}^n frac{a_k}{(1+A_k)^2})=frac{1}{(1+b_{n-1})^2}$$
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!)