An inequality improvement on AMM 11145

Question

I have asked the same question in math.stackexchange, I am reposting it here, looking for answers:

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

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

Answer 1

(extended comment)

Choose some positive numbers $b_1,\dots,b_n$ and denote $a_i=ic_i$ for $i=1,\dots,n$. Use Cauchy-Bunyakovsky-Schwarz inequality $$\left(\sum_{i=1}^k b_ic_i\right)\left(\sum_{i=1}^k b_ic_i^{-1}\right)\geq \left(\sum_{i=1}^k b_i\right)^2$$ to estimate $$ \sum_{k=1}^n \frac{k}{\sum_{i=1}^k b_ic_i}\leq \sum_{k=1}^n \sum_{i=1}^k \frac{k b_i c_i^{-1}}{(b_1+\dots+b_k)^2}=\sum_{i=1}^n c_i^{-1}b_i\sum_{k=i}^n\frac{k}{(b_1+\dots+b_k)^2}. $$ Thus if some positive number $\alpha$ satisfies $$ b_i^2\sum_{k=i}^n\frac{k}{(b_1+\dots+b_k)^2}\leq \alpha,i=1,\dots,n, $$ then we get desired inequality $$ \sum_{k=1}^n \frac{k}{\sum_{i=1}^k a_i}\leq \alpha \sum_{i=1}^n \frac1{a_i}. $$ Choice of $b_i=i$ gives $\alpha=2$ or bit better, but there are rooms for improvement, since for large values $i$ (close to $n$) inequality becomes not very much sharp.

Say, we may try $b_k=k+\lambda$ and optimize by $\lambda$.

Answer 2

An extended comment that has a chance of being useful.

First, note that if you interchange $a_k$ and $a_{k+1}$ then only the $k$-th term on the left side changes and the right side doesn't change at all. From this we can easily see that the worst order for a given set of values is $a_1\le a_2\le\cdots \le a_n$.

Now consider adjusting both $a_k$ and $a_{k+1}$ so that the right side remains unchanged. If $a_k\mapsto a_k+\epsilon$ then $a_{k+1}\mapsto a_{k+1} - \epsilon(a_{k+1}/a_{k})^2+O(\epsilon^2)$. If we are sitting at the worst left side for given right side, then the effect of this adjustment must be $O(\epsilon^2)$ for all $k$. I think that this will give the worst sequence exactly by working from the last term backwards, but I'm out of time just now.

An extension of this approach is to consider adjusting three consecutive terms so that their sum and reciprocal sum both remain the same. Then only two terms of the left side will change so the total effect on the left side can be determined to obtain an explicit condition satisfied by the worst sequence.

I'll start on the last suggestion. First note that for $a,b,c\ge 0$, if we tweak them like this: $$ a\mapsto a+\epsilon, \quad b\mapsto b-\frac{b^2(c^2-a^2)}{a^2(c^2-b^2)}\epsilon, \quad c\mapsto c+\frac{c^2(b^2-a^2)}{a^2(c^2-b^2)}\epsilon $$ then both $x+y+z$ and $1/x+1/y+1/z$ change by only $O(\epsilon^2)$. So now tweak $a_{k-2},a_{k-1},a_k$ like that. Up to $O(\epsilon^2)$, the right side doesn't change and the left side changes by $\Delta_k\epsilon$, where $$ \Delta_k = -\frac{k-2}{S_{k-2}^2} + \frac{k-1}{S_{k-1}^2}\, \frac{a_k^2(a_{k-1}^2-a_{k-2}^2))}{a_{k-2}^2(a_k^2-a_{k-1}^2))}, $$ writing $S_t=\sum_{i=1}^t a_i$. If we are looking at the largest possible left side for given right side, then we must have $\Delta_k=0$ for $3\le k\le n$. This determines $a_3,\ldots,a_n$ in terms of $a_1,a_2$. We can also take $a_1=1$ wlog, so there is one free parameter $a_2$.

Note that for given $a_2$ the sequence is independent of $n$, however for any $a_2>1$, the sequence becomes complex for large enough $n$. The best value of $a_2$ depends on $n$.

Here are numerical values for the maximum value of the left side divided by $\sum_{k=1}^n 1/a_k$, and the approximate value $\hat a_2$ of $a_2$ that achieves it. Also the claimed bound $2-(7 \ln 2)/(8\ln n)$.

n=2, max=1.125, $\hat a_2$=3.000, bnd=1.125
n=3, max=1.205, $\hat a_2$=2.426, bnd=1.448
n=4, max=1.261, $\hat a_2$=2.20, bnd=1.563
n=5, max=1.303, $\hat a_2$=2.07, bnd=1.623
n=10, max=1.425 $\hat a_2$=1.83, bnd=1.737
n=100, max=1.6844 $\hat a_2$=1.5688, bnd=1.868

These values suggest that the claimed bound is correct, but far from sharp except for $n=2$.

Here is the optimal sequence for $n=100$: 1.0, 1.56881, 2.10179, 2.62862, 3.15987, 3.70074, 4.25434, 4.82275, 5.40752, 6.00993, 6.63103, 7.27180, 7.93314, 8.61591, 9.32096, 10.0491, 10.8013, 11.5783, 12.3810, 13.2103, 14.0673, 14.9527, 15.8678, 16.8135, 17.7909, 18.8012, 19.8456, 20.9254, 22.0419, 23.1965, 24.3906, 25.6258, 26.9037, 28.2261, 29.5947, 31.0114, 32.4783, 33.9974, 35.5710, 37.2016, 38.8915, 40.6436, 42.4605, 44.3455, 46.3016, 48.3323, 50.4413, 52.6325, 54.9100, 57.2782, 59.7421, 62.3066, 64.9774, 67.7603, 70.6618, 73.6887, 76.8485, 80.1493, 83.5998, 87.2096, 90.9890, 94.9493, 99.1028, 103.463, 108.045, 112.865, 117.942, 123.294, 128.945, 134.919, 141.244, 147.950, 155.074, 162.653, 170.734, 179.367, 188.610, 198.531, 209.209, 220.733, 233.213, 246.773, 261.566, 277.773, 295.617, 315.370, 337.372, 362.055, 389.972, 421.852, 458.671, 501.779, 553.098, 615.480, 693.382, 794.237, 931.624, 1133.84, 1473.84, 2238.74

Answer 3

(This doesn't answer the actual question, but it was too long for a comment.Besides, it's related, and someone might even find it relevant.)

The following attempts $(1)$ to prove a non-discrete counterpart using integrals to the AMM 11145 inequality for an arbitrary real function $a(x)$ that is integrable and strictly positive on $[0, n]$: $$ \int_0^n \frac{k}{\int_0^k a(x)\,dx}\,dk\leq 2\int_0^n \frac{1}{a(k)}\,dk $$ and also $(2)$ prove that $2$ is the smallest constant for which it holds.

The notations are to emphasize the relationship to the discrete case, but $n, k, x, a(x)$ are all reals, of course. An equivalent integral inequality for $[0,\infty)$ was mentioned by @1015 here https://math.stackexchange.com/questions/599999/the-series-sum-limits-n-1-infty-frac-n-frac1a-1-frac1a-2-dotsb-fra/600943#600943 but was not elaborated or further pursued at the time.

$(1)$ The Cauchy-Schwartz inequality for real-valued functions

$$ \left\lvert{\int_0^k f(x)\,g(x)\,dx}\right\rvert^2 \le \int_0^k f(x)^2 \,dx \int_0^k g(x)^2 \,dx $$ written for $f(x) = x / {\sqrt{(a(x))}}$ and $g(x) = \sqrt{(a(x))}$ gives

$$ \left\lvert{\int_0^k{x\,dx}}\right\rvert^2 \le \int_0^k \frac{x^2}{a(x)}\,dx\;\int_0^k a(x)\,dx $$

then replacing the left hand side with the obvious $k^4 / 4$ and rearranging

$$ \frac{k}{\int_0^k a(x)\,dx} \le \frac{4}{k^3}\;\int_0^k\frac{x^2}{a(x)}\,dx $$

Integrating in $k$ over $[0, n]$ then applying Fubini's theorem and simplifying gives

$$ \begin{align} \int_0^n \frac {k} {\int_0^k a(x)\,dx}\,dk & \le \int_0^n \int_0^k\frac{4}{k^3}\; \frac{x^2}{a(x)}\,dk\,dx \\ & = \int_0^n \int_x^n\frac{4\,x^2}{a(x)}\; \frac{1}{k^3}\,dx\,dk \\ & = \int_0^n \frac{4\,x^2}{a(x)}\; (\frac{1}{2\,x^2} - \frac{1}{2\,n^2})\,dx \\ & = 2\;\int_0^n \frac{1}{a(x)}\; (1 - \frac{x^2}{n^2})\,dx \\ & \le 2\;\int_0^n \frac{1}{a(x)}\,dx = 2\;\int_0^n \frac{1}{a(k)}\,dk \end{align} $$

which proves the claimed inequality.

$(2)$ To prove that $C = 2$ is the best constant, consider the function $a(x) = x + \epsilon$. After calculating the elementary integrals, the inequality reduces to:

$$ 2\,(\,ln(n + 2\,\epsilon) - ln(2\,\epsilon)\,) \le 2\,(\,ln(n + \epsilon) - ln(\epsilon)) $$

The ratio of the two sides approaches $1$ as $\epsilon \to 0$ as can be easily verified using l'Hopital rule for example:

$$ \begin{align} \lim_{\epsilon\to 0}{ln(n + 2\,\epsilon) - ln(2\,\epsilon) \over ln(n + \epsilon) - ln(\epsilon)} = {\frac{2}{n+2\,\epsilon} - \frac{2}{2\,\epsilon} \over \frac{1}{n+\epsilon} - \frac{1}{\epsilon}} = {\frac{-\,n}{n + 2\,\epsilon} \over \frac{-\,n}{n+\epsilon}} = 1 \end{align} $$

...which indicates that any constant $C' < 2$ would fail to satisfy the inequality for small enough $\epsilon$.

To sum it up, the result itself is not surprising. The proofs for the discrete case inequality usually showed that $2$ is the best constant for arbitrarily large series as well. The formulation using integrals just allows for the equivalent of such infinite series to occur within a finite $[0,n]$ interval. It might be also worth noting that the linear function used in the second step reminisces of the proofs by harmonic numbers in the discrete case (that $2$ is optimal for infinite series).

Answer 4

I came up with something years ago which is similar to Fedor Petrov's (or other users').

Problem: Let $a_i > 0; \ i = 1, 2, \cdots, n$ ($n\ge 2$). Let $C(n) = 2 - \frac{7\ln 2}{8\ln n}$. Prove that $$\sum_{k=1}^n \frac{k}{a_1 + a_2 + \cdots + a_k} \le C(n)\sum_{k=1}^n \frac{1}{a_k}.$$

Introducing the coefficients (to be determined) $C_1, C_2, \cdots, C_n > 0$, by the Cauchy-Bunyakovsky-Schwarz inequality, we have $$\frac{k}{a_1 + a_2 + \cdots + a_k} \le \frac{k}{(C_1+C_2 + \cdots + C_n)^2}\sum_{i=1}^k \frac{C_i^2}{a_i}$$ and \begin{align} \sum_{k=1}^n \frac{k}{a_1 + a_2 + \cdots + a_k} &\le \sum_{k=1}^n \frac{k}{(C_1+C_2 + \cdots + C_n)^2}\sum_{i=1}^k \frac{C_i^2}{a_i}\\ &= \sum_{k=1}^n \Big[C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2}\Big]\frac{1}{a_k}. \end{align} Equality occurs if and only if $C_k = a_k, \ k=1, 2, \cdots, n$.

The problem becomes: Can we choose $C_k > 0,\ k=1, 2, \cdots, n$ such that $$\sup_{k=1, 2, \cdots, n} C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2} \le C(n), \ \forall n\ge 2 ?$$

It is easy if $C(n)=2$. For a simple proof, I chose $$C_k = \sqrt{k(k+1)(k+2)(k+3)} - \sqrt{(k-1)k(k+1)(k+2)}, \ k = 1, 2, \cdots, n.$$ We have $C_1 + C_2 + \cdots + C_m = \sqrt{m(m+1)(m+2)(m+3)}$ and \begin{align} &C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2}\\ = \ &k(k+1)(k+2)\big(2k+2 - 2\sqrt{(k+3)(k-1)}\big)\\ &\quad \cdot\Big(\frac{1}{2(k+1)} - \frac{1}{2(k+2)} - \frac{1}{2(n+2)(n+3)}\Big)\\ \le \ & k(k+1)(k+2)\big(2k+2 - 2\sqrt{(k+3)(k-1)}\big) \Big(\frac{1}{2(k+1)} - \frac{1}{2(k+2)}\Big)\\ \le \ & 2. \end{align}

Fedor Petrov chose $C_k = k,\ \forall k$. We have $$C_k^2\sum_{m=k}^n \frac{m}{(C_1+C_2+\cdots+C_m)^2} = k^2\Big(-\frac{4}{n+1}+4\Psi(1, n+2)+\frac{4}{k}-4\Psi(1, 1+k)\Big).$$ Denote the RHS as $f(k, n)$. We have the asymptotic expansion $f(\sqrt{n}, n) \sim 2 - \frac{2}{3\sqrt{n}} - \frac{2}{n} + \cdots (n\to \infty)$. Thus, it at most gives $C(n) \ge 2 - \frac{A}{\sqrt{n}}$ where $A$ is a constant.

Fedor Petrov also suggested $C_k = k + \lambda$ (optimize $\lambda$). If $\lambda$ is a constant, similarly, we have $f(\sqrt{n},n, \lambda) \sim 2 - \frac{4\lambda + 2}{3\sqrt{n}} - \cdots (n \to \infty)$. It is not enough.