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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.


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