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# Problem A. 709. (November 2017)

A. 709. Let $\displaystyle a>0$ be a real number. Find the minimal constant $\displaystyle C_a$ for which the inequality

$\displaystyle C_a\sum_{k=1}^n \frac1{x_k-x_{k-1}} >\sum_{k=1}^n \frac{k+a}{x_k}$

holds for any positive integer $\displaystyle n$ and any sequence $\displaystyle 0=x_0<x_1<\cdots<x_n$ of real numbers.

(5 pont)

Deadline expired on December 11, 2017.

Answer. If $\displaystyle a>\frac12$, then $\displaystyle C_a=\frac{(2a+1)^2}{4a}$. If $\displaystyle a\le \frac12$, then $\displaystyle C_a=2$.

Solution. Using Titu's lemma for $\displaystyle k=1,2,\dots,n-1$:

$\displaystyle \frac{k^2}{x_k}+\frac{(2a+1)^2}{x_{k+1}-x_k}\ge \frac{(k+2a+1)^2}{x_{k+1}},$

where equality holds iff $\displaystyle \frac{x_k}{k}=\frac{x_{k+1}-x_k}{2a+1}$, i.e. $\displaystyle \frac{x_{k+1}}{x_k}=\frac{k+(2a+1)}{k}$. Adding all these up:

$\displaystyle (2a+1)^2\left(\frac1{x_2-x_1}+\dots+\frac1{x_n-x_{n-1}}\right)\ge -\frac1{x_1}+\frac{n^2}{x_n}+\sum_{k=1}^{n-1} \frac{(k+2a+1)^2-(k+1)^2}{x_{k+1}},$

which rearranges to

 $\displaystyle \frac{(2a+1)^2}{4a}\left(\frac1{x_1}+\frac1{x_2-x_1}+\dots+\frac1{x_n-x_{n-1}}\right)\ge \frac{n^2}{4ax_n}+\sum_{k=1}^n \frac{k+a}{x_k},$ $\displaystyle (\star)$

hence the constant $\displaystyle C_a=\frac{(2a+1)^2}{4a}$ works for all $\displaystyle a>0$. However, we can do better for $\displaystyle a\le \frac12$, as then

$\displaystyle \frac{(2\cdot (1/2)+1)^2}{4\cdot (1/2)}\left(\sum_{k=1}^n \frac1{x_k-x_{k-1}}\right)>\sum_{k=1}^n \frac{k+1/2}{x_k}\ge \sum_{k=1}^n \frac{k+a}{x_k},$

so $\displaystyle C_a=\frac{(2\cdot (1/2)+1)^2}{4\cdot (1/2)}=2$ also works.

Next, we show that these constants, i.e. $\displaystyle C_a=\frac{(2a+1)^2}{4a}$ for $\displaystyle a>\frac12$ and $\displaystyle C_a=2$ for $\displaystyle a\le \frac12$, are optimal. To this end, first note a technical lemma:

Lemma. For any $\displaystyle b>0$, there exist constants $\displaystyle 0<c_1<c_2$ such that $\displaystyle \forall N$

$\displaystyle c_1N^b<\prod_{j=1}^N \left(1+\frac{b}{j}\right)<c_2N^b.$

Proof. We use that $\displaystyle 1+x\approx e^x$. This is made precise by the known estimate $\displaystyle e^{\frac{x}{x+1}}\le 1+x\le e^x$:

$\displaystyle \exp\left(\sum_{j=1}^N \frac{b}{b+j}\right)\le \prod_{j=1}^N \left(1+\frac{b}{j}\right)\le \exp\left(\sum_{j=1}^N \frac{b}{j}\right).$

From the known integral estimate, we have $\displaystyle \sum_{j=1}^N \frac1{c+j}\sim \log N$, hence the lemma. (The lemma also holds for $\displaystyle b\le 0$.) $\displaystyle \blacksquare$

Case 1: $\displaystyle a>\frac12$. Let $\displaystyle x_1=1$ and $\displaystyle x_k=\prod_{j=1}^{k-1}\frac{j+(2a+1)}{j}$ ($\displaystyle k=2,3,\dots,n,\dots$). We claim that then as $\displaystyle n\to\infty$,

$\displaystyle \left(\sum_{k=1}^n \frac{k+a}{x_k}\right)/\left(\sum_{k=1}^n \frac1{x_k-x_{k-1}}\right)\to \frac{(2a+1)^2}{4a},$

and so no constant less than $\displaystyle \frac{(2a+1)^2}{4a}$ could work. Since we chose sequence $\displaystyle (x_k)$ such that equality holds at each application of Titu's lemma, equality holds in $\displaystyle (\star)$, and therefore it suffices to show that as $\displaystyle n\to\infty$,

 $\displaystyle \frac{n^2}{4ax_n}/\left(\sum_{k=1}^n \frac1{x_k-x_{k-1}}\right)\to 0.$ $\displaystyle (\dagger)$

Since for $\displaystyle k\ge 2$

$\displaystyle x_k-x_{k-1}=\prod_{j=1}^{k-1}\frac{j+(2a+1)}{j}-\prod_{j=1}^{k-2}\frac{j+(2a+1)}{j}=\frac{2a+1}{k-1}\prod_{j=1}^{k-2}\frac{j+(2a+1)}{j}=\prod_{i=1}^{k-1}\frac{i+2a}{i},$

the lemma tells us that for some constants $\displaystyle 0<c_1<c_2$ and $\displaystyle 0<c_3<c_4$, $\displaystyle c_1N^{2a+1}<x_N<c_2N^{2a+1}$ and $\displaystyle c_3N^{2a}<x_N-x_{N-1}<c_4N^{2a}$. As is well-known (e.g. by integral estimate of the zeta function), the latter implies (if $\displaystyle a>\frac12$) that $\displaystyle \sum_{k=1}^\infty \frac1{x_k-x_{k-1}}$ is convergent. From this, $\displaystyle (\dagger)$ is clear.

Case 2: $\displaystyle a\le \frac12$. Simply let $\displaystyle x_k=k^2$ for each $\displaystyle k$. (This rhymes with $\displaystyle x_N=\Theta(N^{2a+1})$ from the previous case, where $\displaystyle \Theta$ is Big O notation.) Then

$\displaystyle \sum_{k=1}^n \frac1{x_k-x_{k-1}}=\sum_{k=1}^n \frac1{2k-1}\sim \frac12\log n,$

$\displaystyle \sum_{k=1}^n \frac{k+a}{x_k}=\sum_{k=1}^n\frac1k+a\sum_{k=1}^n \frac1{k^2}\sim \log n,$

and thus

$\displaystyle \left( \sum_{k=1}^n \frac{k+a}{x_k}\right)/\left(\sum_{k=1}^n \frac1{x_k-x_{k-1}}\right)\to 2,$

proving the optimality of $\displaystyle C_a=2$. This completes the proof in both cases.

### Statistics:

 6 students sent a solution. 5 points: Schrettner Jakab. 2 points: 2 students. 1 point: 1 student. 0 point: 2 students.

Problems in Mathematics of KöMaL, November 2017