Let $y_1 = x_1-x_0$, $y_2 = x_2-x_1$, \dots, $y_n = x_n-x_{n-1}$. So $x_k = y_1 + \dots + y_n$, meaning the inequality reads \[ \frac{1}{y_1} + \dots + \frac{1}{y_n} \ge a \left( \frac{2}{y_1} + \frac{3}{y_1+y_2} + \dots + \frac{n+1}{y_1+\dots+y_n} \right). \] We claim the optimal constant is $a = 4/9$. First we show this is sharp as $n \to \infty$ by taking $y_k = k(k+1)$ for each $k$, thus $x_k = \frac{1}{3} k(k+1)(k+2)$. Then the inequality telescopes as \[ 1 - \frac{1}{n+1} \ge a \left( \frac{3}{2} \left( \frac11+\frac12 -\frac{1}{n+1}-\frac{1}{n+2} \right) \right) \]which when $n \to \infty$ forces $a \le \frac49$. To prove the inequality itself, we use Cauchy: \begin{align*} \frac{9}{y_1} &= \frac{9}{y_1} \\ \frac{1}{y_1} + \frac{9}{y_2} &\ge \frac{16}{y_1+y_2} \\ \frac{4}{y_1+y_2} + \frac{9}{y_3} &\ge \frac{25}{y_1+y_2+y_3} \\ \frac{9}{y_1+y_2+_3} + \frac{9}{y_4} &\ge \frac{36}{y_1+y_2+y_3+y_4} \\ \frac{16}{y_1+y_2+y_3+y_4} + \frac{9}{y_5} &\ge \frac{49}{y_1+y_2+y_3+y_4+y_5} \\ &\vdots \\ \frac{(n-1)^2}{y_1+\dots+y_{n-1}} + \frac{9}{y_n} &\ge \frac{(n+2)^2}{y_1+\dots+y_n} \end{align*}which when summed over $k$ telescopes to the right thing. Remark: It's interesting to tabulate the answers for $n$. If one takes $n=1$, then we get $a \le \frac{1}{2}$, and when $n=2$ one finds the optimal solution is $(y_1, y_2) = (1,5)$ for $a \le \frac{12}{25} = 0.48$. It is hard to solve $n=3$, but one can get things less than $0.48$ without too much difficulty. So this gives some sense for where the constant is around. Remark: One way to guess answer is that there's not many sums of fractions of this shape that we can add up (since small cases make it clear that $y_i$ should grow quickly), so best guess is telescoping sum $y_k = k(k+1)$, which nicely cancels stuff on the right hand side as we then get $x_k = \frac 13 k(k+1)(k+2)$, so $\frac{k+1}{x_k} = \frac{3}{k(k+2)}$ also telescopes. A better motivation is to start with the Titu lemma as above and then try to engineer constants to match the problem (there are enough degrees of freedom that this ``must'' be possible). I find this a plausible way the problem was written.

As I know this problem was discarded from the Shortlist as known. Does someone know the source?

This problem might become easy if the largest $a$ is known. However, it seems that it's not easy to guess the value of $a$ (though one may know that $a<0.48$ and is around 0.48). Here's my motivation (together with my solution) for guessing $a_{max} = 4/9$. As above, let $y_i = x_i-x_{i-1}$. Fix $n$ for a while. It's reasonable to apply Cauchy's here. To make sure that the equality holds, suppose that when $y_i = b_i^2$, $a$ achieves its maximum (note that $n$ is fixed now). Then by Cauchy's, \[\sum_{k=1}^{i} \frac{b_k^2}{y_k}\geq \frac{(\sum_{k=1}^{i}b_k)^2}{\sum_{k=1}^{i}y_k}......(i\textup{-th inequality})\] Now, multiply the $i$-th inequality by $\frac{1}{b_i^2}-\frac{1}{b_{i+1}^2}$ (here $b_{n+1} = \infty$) and sum them all: \[\sum_{i=1}^{n}\frac{1}{y_i}\geq \sum_{i=1}^{n}(\frac{1}{b_i^2}-\frac{1}{b_{i+1}^2})\frac{(\sum_{k=1}^{i}b_k)^2}{\sum_{k=1}^{i}y_k}......(\Delta)\] as long as $b_i$ is non-decreasing. Therefore, we hope that $(\frac{1}{b_i^2}-\frac{1}{b_{i+1}^2})(\sum_{k=1}^{i}b_k)^2 = (i+1)a$. At this point, $b_{n+1}=\infty$ would be annoying, so it's reasonable to find instead a non-decreasing infinite sequence $b_i$ satisfying \[(\frac{1}{b_i^2}-\frac{1}{b_{i+1}^2})(\sum_{k=1}^{i}b_k)^2 = (i+1)a\quad\forall i\in\mathbb{N}......(\circ)\] where $a$ is a positive constant. Then in order to get $(\Delta)$, we have to add $\frac{b_n^2}{b_{n+1}^2 y_i}\geq 0$ to it. Here we make a relaxation that is not sharp, so we hope that the difference tends to zero as $n$ tends to infinity. In other words, $b_i$ should be unbounded. To simplify $(\circ)$, let $x_i = \frac{\sum_{k=1}^{i}b_k}{b_i}$. Then $x_i^2-(x_{i+1}-1)^2 = (i+1)a$. It's natural to guess that $x_i$ is a linear function w.r.t. $i$ (or first guess that $x_i$ is a polynomial and then deduce that it must be a linear function). With the constraint that $x_1=1$, it's easy to show that $x_i = \frac{1}{3}i+\frac{2}{3}$. Hence, $a=\frac{4}{9}$ and \[b_i = b_1\prod_{k=1}^{i-1}\frac{b_{k+1}}{b_k} = b_1\prod_{k=1}^{i-1}\frac{x_k}{x_{k+1}-1} = b_1\prod_{k=1}^{i-1}\frac{k+2}{k} = \frac{i(i+1)}{2}b_1\] Note that $b_i$ is non-decreasing and unbounded, so $a=\frac{4}{9}$ is the largest number that satisfies the condition.

silouan wrote: As I know this problem was discarded from the Shortlist as known. Does someone know the source? Because it was similar to one of Japan MO problem. Japan observers show a pretty similar problem (and also steps to solve it).

The main difficulty in this problem is definitely obtaining the constant. Unfortunately, it was not natural for me to guess the form of the coefficients. I will be interested to know if there are any other hints that $a=\frac{4}{9}$, but for now @USJL's motivation might be the best we have. But after obtaining or guessing the exact value of $a$, it is fairly straightforward to obtain the equality case by derivatives, and we simply match the corresponding weights after that on the respective Cauchy inequalities.

I don't know which Japan MO problem is referenced in post #5, but a similar problem appeared in the American Mathematical Monthly in 2005: American Mathematical Monthly Problem 11145 wrote: Determine the least real number $c$, such that for any integer $n\geq 1$ and any positive real numbers $a_1, a_2, \ldots a_n$ the following holds: \[\sum_{k=1}^{n}\frac{k}{\frac{1}{a_1}+\frac{1}{a_2}+\dotsb +\frac{1}{a_k}} < c\sum_{k=1}^{n}a_k.\] If you really enjoy computation, try the following generalization of both problems: Let $t$ be a real number. Find the smallest real number $c$ such that the following inequality holds for all positive integers $n$ and positive real numbers $a_1, \dotsc, a_n$: \[\frac{1+t}{a_1}+\frac{2+t}{a_1+a_2}+\dotsb+\frac{n+t}{a_1+\dotsb+a_n} \leq c\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_n}\right).\]

a1267ab wrote: Let $t$ be a real number. Find the smallest real number $c$ such that the following inequality holds for all positive integers $n$ and positive real numbers $a_1, \dotsc, a_n$: \[\frac{1+t}{a_1}+\frac{2+t}{a_1+a_2}+\dotsb+\frac{n+t}{a_1+\dotsb+a_n} \leq c\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_n}\right).\] This generalisation was proposed as problem A.709 in KöMaL. See here.

a1267ab wrote: I don't know which Japan MO problem is referenced in post #5, but a similar problem appeared in the American Mathematical Monthly in 2005: (...) KöMaL N.189. Prove that arbitrary positive integers $a_1, ..., a_n$ satisfy the following inequality: $$ a_1 + \frac{2}{\frac1{a_1}+\frac1{a_2}} + \frac{3}{\frac1{a_1}+\frac1{a_2}+\frac1{a_3}} +\ldots + \frac{n}{\frac1{a_1}+\frac1{a_2}+\ldots+\frac1{a_n}} < 2(a_1+a_2+\ldots+a_n). $$(November, 1998)

silouan wrote: As I know this problem was discarded from the Shortlist as known. Does someone know the source? When I was preparing for IMO in 2012, I was given this problem with a=1/4 and asked to prove the strict inequality. Then, I came up with a proof similar to that of Evan’s which my instructor worked on and showed the technique can be used to prove a stronger constant, a=4/9. Thats my last experience with the problem until when I just saw it on the shortlist today and it rang a lot of bells. I’m glad it was discarded. Best, Alyazeed

We claim that the constant is $\boxed{a = \frac{4}{9}}$. The following proof proceeds in two steps: Let $x_i = a_1 + a_2 + \dots + a_i$. By the hypothesis of the problem, we have $a_1 , a_2, \dots \in \mathbb{R}^+$ Then the problem transforms to \[ \sum_{i = 1}^n \frac{1}{a_i} > k \cdot \left( \sum_{i = 1}^n \frac{i + 1}{\sum_{j = 1}^{i} a_j } \right) \]$\textbf{Claim 01.}$ Any constant $a > \frac{4}{9}$ fails. $\textit{Proof.}$ Consider the sequence $a_i = \frac{i(i+1)}{6}$ for $i \in \mathbb{N}$. Therefore, \[ \sum_{i = 1}^n \frac{6}{i(i+1)} > a \cdot \left( \sum_{i = 1}^n \frac{18}{i(i+2)} \right) \]\[ a < \frac{2}{3} \cdot \frac{1 - \frac{1}{n}}{\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}} \]for all $n \in \mathbb{N}$. Hence, we have \[ a \le \lim_{n \to \infty} \frac{2}{3} \cdot \frac{1 - \frac{1}{n}}{\frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2}}= \frac{4}{9} \]$\textbf{Claim 02.}$ The constant $a = \frac{4}{9}$ works. $\textit{Proof.}$ Notice that by CS Engel, \[ \frac{\frac{(i-1)^2}{9}}{a_1 + a_2 + \dots + a_{i - 1}} + \frac{1}{a_i} \ge \frac{ \frac{(i + 2)^2}{9}}{a_1 + a_2 + \dots + a_i} \]Therefore, summing all such inequalities gives us \[ \frac{1}{9a_1} + \sum_{i = 2}^n \frac{1}{a_i} \ge \frac{4}{9} \cdot \sum_{i = 2}^n \frac{i + 1}{a_1 + a_2 + \dots + a_i} \]Hence, we have \[ \sum_{i = 1}^n \frac{1}{a_i} \ge \frac{4}{9} \cdot \sum_{i = 1}^n \frac{i + 1}{a_1 + a_2 + \dots + a_i} \] Remark. I'm quite stupid to bash the case $n = 3$, which results $a$ being an ugly irrational. Then I get $a = \frac{12}{25}$ for $k = 2$ and $a = \frac{1}{2}$ for $k = 1$. To easily apply CS Engel, i expect to have $k = \frac{1 - n^2}{2}$ for $n \in \mathbb{Q}$. To prove $a > \frac{4}{9}$ fails, consider the equality case of when $a = \frac{4}{9}$. Indeed, I spent 3 hours just to find the right constant.

Remark: I found the bound first?? Naturally let $t_i = x_i - x_{i - 1}$ for all $i : 1 \to n$ with $t_1 = x_1 - x_0 = x_1$. The inequality becomes\[\sum_{i = 1}^n \frac{1}{t_i} \geq a \left(\sum_{i = 1}^n\frac{i + 1}{t_1 + \ldots + t_i}\right).\]Since the denominators of terms in the RHS are the partial sums of $x_i$'s we are motivated to find some bound by Titu's. Indeed, after some time, we arrive at\[\frac{m^2}{t_1 + \ldots + t_m} + \frac{9}{t_{m + 1}} \geq \frac{(m + 3)^2}{t_1 + \ldots + t_{m + 1}}\]for all positive integers $m$ from $0 \to n - 1$. This is very good: upon summing over all terms, we get that\[9\left(\sum_{i = 1}^n \frac{1}{t_i}\right) + \sum_{i = 1}^{n-1} \frac{i^2}{t_1 + \ldots + t_i} \geq \sum_{i = 1}^n \frac{(i + 2)^2}{t_1 + \ldots + t_i} \implies 9\left(\sum_{i = 1}^n \frac{1}{t_i}\right) \geq 4\left(\sum_{i = 1}^{n} \frac{i + 1}{t_1 + \ldots + t_i}\right) \]showing that the constant $a = \tfrac49$ works. However, it is not yet clear that $\tfrac49$ is the largest such constant. In order for equality conditions to hold for all inequalities above, we need\[\frac{m^2}{9} = \frac{(t_1 + \ldots + t_m)^2}{t_{m + 1}^2}\]for all $m : 1 \to n - 1$. Indeed, it is not particularly hard to verify that the sequence $t_i = \tbinom{i + 1}{2}$ verifies these equality conditions, motivated by letting $t_1 = 1$. Lastly, we plug this sequence back into the original inequality in $t_i$'s. By telescoping, it is clear that\[\sum_{i = 1}^n \frac{1}{t_i} = 2\left(1 - \frac{1}{n + 1}\right)\]and\[\sum_{i = 1}^n \frac{i + 1}{t_1 + \ldots t_i} = \sum_{i = 1}^n \frac{i + 1}{\binom{i + 2}{3}} = \sum_{i = 1}^n \frac{6}{i(i + 2)} = 3\left(\frac{3}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}\right)\]and as $n$ tends to infinity the ratio of these two does indeed approach $2/(9/2) = \tfrac49$ hence our desired answer is $a = \tfrac49$. $\blacksquare$

Similar to Japan TST 2016-7. Problem. Determine the smallest real number $k$ such that, for all integer $n\geq 2$ and for all positive real numbers $a_1,a_2,\dots,a_n$ we have $$\displaystyle\frac{1}{a_1+a_2}+\frac{1}{a_1+a_2+a_3}+\cdots+\frac{1}{a_1+a_2+\cdots+a_n} < k\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right).$$

It's easy to guess that answer

and we can prove it by C-S.

Solved with Rama1728 Claim 1: If $a>4/9$ then the ineq fails. Proof: Let $x_n=\frac{n(n+1)(n+2)...(n+i-1)}{i}$ where $i$ is a positive integer greater than $2$. Then its easy to note that we have $x_n-x_{n-1}=n(n+1)(n+2)...(n+i-2)$, now we will put 2 great results that are left to the reader: $$\sum_{k=1}^{n} \frac{1}{x_k-x_{k-1}}=\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)...(k+i-2)}=\frac{1}{i-2} \left(\frac{1}{(i-2)!}-\frac{1}{(n+1)(n+2)...(n+i-2)} \right)$$$$\sum_{k=1}^{n} \frac{k+1}{x_k}=\sum_{k=1}^{n} \frac{i}{k(k+1)(k+2)...(k+i-1)}+\frac{i}{(k+1)(k+2)...(k+i-1)}=\frac{1}{(i-1)(i-2)} \left(\frac{2i^2-3i}{(i-1)!}-\frac{(n-2)i^2-n+1}{(n+1)(n+2)...(n+i-1)} \right)$$Now note that when $n \to \infty$ the ineq becomes: $$\frac{1}{(i-2)!} \ge \frac{a(2i^2-3i)}{(i-1)(i-1)!} \implies a \le \frac{i^2-2i+1}{2i^2-3i} \overset{i=3}{\implies} a \le \frac{4}{9}$$Thus our claim is proven!. Main solution: We need to show that $a=\frac{4}{9}$ is the maximun, now note that by titu's lemma we have: $$\sum_{k=1}^{n} \frac{9}{x_k-x_{k-1}}+\sum_{k=2}^{n} \frac{(k-1)^2}{x_{k-1}} \ge \sum_{k=1}^{n} \frac{(k+2)^2}{x_k} \implies \sum_{k=1}^{n} \frac{9}{x_k-x_{k-1}} \ge \sum_{k=1}^{n} \frac{4k+4}{x_k}+\frac{n^2}{x_n} \ge \sum_{k=1}^{n} \frac{4k+4}{x_k}$$Thus we are done

MathLuis wrote: Solved with Rama1728 Claim 1: If $a>4/9$ then the ineq fails. Proof: Let $x_n=\frac{n(n+1)(n+2)...(n+i-1)}{i}$ where $i$ is a positive integer greater than $2$. Then its easy to note that we have $x_n-x_{n-1}=n(n+1)(n+2)...(n+i-2)$, now we will put 2 great results that are left to the reader: $$\sum_{k=1}^{n} \frac{1}{x_k-x_{k-1}}=\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)...(k+i-2)}=\frac{1}{i-2} \left(\frac{1}{(i-2)!}-\frac{1}{(n+1)(n+2)...(n+i-2)} \right)$$$$\sum_{k=1}^{n} \frac{k+1}{x_k}=\sum_{k=1}^{n} \frac{i}{k(k+1)(k+2)...(k+i-1)}+\frac{i}{(k+1)(k+2)...(k+i-1)}=\frac{1}{(i-1)(i-2)} \left(\frac{2i^2-3i}{(i-1)!}-\frac{(n-2)i^2-n+1}{(n+1)(n+2)...(n+i-1)} \right)$$Now note that when $n \to \infty$ the ineq becomes: $$\frac{1}{(i-2)!} \ge \frac{a(2i^2-3i)}{(i-1)(i-1)!} \implies a \le \frac{i^2-2i+1}{2i^2-3i} \overset{i=3}{\implies} a \le \frac{4}{9}$$Thus our claim is proven!. Main solution: We need to show that $a=\frac{4}{9}$ is the maximun, now note that by titu's lemma we have: $$\sum_{k=1}^{n} \frac{9}{x_k-x_{k-1}}+\sum_{k=2}^{n} \frac{(k-1)^2}{x_{k-1}} \ge \sum_{k=1}^{n} \frac{(k+2)^2}{x_k} \implies \sum_{k=1}^{n} \frac{9}{x_k-x_{k-1}} \ge \sum_{k=1}^{n} \frac{4k+4}{x_k}+\frac{n^2}{x_n} \ge \sum_{k=1}^{n} \frac{4k+4}{x_k}$$Thus we are done I'd like to add in some motivational remarks: First of all, to get a bound over \(a\), we need to find a suitable sequence \(x_i\). First we thought about the A.P sequence, because the LHS will have common denominator, but the RHS is disgusting. It then reminded us about telescoping sums. What can we substitute in order to make \(x_i-x_{i-1}\) and \(\frac{i}{x_{i-1}}\) to be a telescope-able term? Then we thought for a while and tried the substitution @MathLuis had added, and it worked. So the main idea for the disproval of \(a>4/9\) is telescoping sums. Now to prove the other part of the inequality, T2 is natural, because of the denominators in the LHS. Now we send \(9\) to that side and \(4\) remains on the RHS. Now, we want to apply T2 in such a way that we get \(4k+4\) in the numerator of each term in the RHS. So, we want \((k+a)^2-(k+b)^2=4k+4\) and \(a+3=b\) and a simple check gives \(a=-1,b=2\). From here, everything is pretty straightforward

The answer is $a=\frac 49$. The construction is $d_i=x_i-x_{i-1}=\binom{i+2}{2}$. We can check that it works. It is inspired by binomial coefficients for ease of computation. The bound: Inspired by the equality case, observe by Cauchy-Schwartz, $$\frac{1}{d_1+\cdots+d_n} = \frac{1}{\binom 22 \frac{d_1}{\binom 22} + \binom 32 \frac{d_2}{\binom 32}+\cdots+\binom{n+1}{2} \frac{d_n}{\binom{n+1}{2}}}\le \left(\frac{1}{\sum\limits_{j=2}^{n+1} \binom j2} \right)^2 \left(\sum\limits_{j=2}^{n+1} \frac{\binom j2}{d_{j-1}/\binom j2}\right) $$ $$=\binom{n+2}{3}^{-2} \sum\limits_{j=2}^{n+1} \frac{\binom j2^2}{d_{j-1}}$$ Hence it suffices to show that $$\sum\limits_{j=1}^n \frac{1}{d_j} - a\sum\limits_{j=1}^n \frac{j+1}{d_1+\cdots+d_j}\ge \sum\limits_{j=1}^n \frac{1}{d_j} - a\sum\limits_{j=1}^n (j+1)\binom{j+2}{3}^{-2} \sum\limits_{k=2}^{j+1} \frac{\binom k2^2}{d_{k-1}}$$ Matching the $\frac{1}{d_j}$ coefficient, it suffices to show $$1-\frac 49 \binom{j+1}{2}^2 \sum\limits_{k\ge j} \frac{k+1}{\binom{k+2}{3}^2} \ge 0 \iff \sum\limits_{k\ge j} \frac{1}{(k+2)^2(k+1)k^2} \le \frac{1}{4j^2(j+1)^2}$$ We can see $$\sum\limits_{k\ge j} \frac{1}{(k+2)^2 (k+1) k^2} = \sum\limits_{k\ge j} \frac{1}{(k+2)k} \frac{1}{(k+2)(k+1)k} = \sum\limits_{k\ge j} \left(\frac{1/2}{k} - \frac{1/2}{k+2} \right) \left( \frac{1/2}{k} - \frac{1}{k+1} + \frac{1/2}{k+2} \right) $$ $$= \sum\limits_{k\ge j} \left(\frac{1/4}{k^2} - \frac{1/4}{(k+2)^2} - \frac{1/2}{k(k+1)} + \frac{1/2}{(k+1)(k+2)} \right)= \frac{1/4}{j^2} + \frac{1/4}{(j+1)^2} - \frac{1/2}{j(j+1)}= (\frac{1/2}{j}-\frac{1/2}{j+1})^2 = \frac{1}{2j(j+1)}^2 $$

Let $d_i=x_i-x_{i-1}$ for positive integers $i\le n$. Note that $d_i>0$ for all $i$. We claim that \[\frac{1}{d_1} + \frac{1}{d_2} + \dots + \frac{1}{d_n} \geq \frac49 \left( \frac{2}{x_1} + \frac{3}{x_2} + \dots + \frac{n+1}{x_n} \right)\]Observe that by Titu's Lemma, for $i\ge 2$, $\tfrac{9}{d_i}+\tfrac{(i-1)^2}{x_{i-1}}\ge \tfrac{(i+2)^2}{x_i}$ so $\tfrac{1}{d_i}\ge \tfrac{(i+2)^2}{9x_i}-\tfrac{(i-1)^2}{9x_{i-1}}$. We have \begin{align*} \sum_{i=1}^{n}{\frac1{d_i}}&=\frac{1}{x_1}+\sum_{i=2}^{n}{\frac1{d_i}}\\ &\ge \frac{1}{x_1}+\sum_{i=2}^{n}{\left(\frac{(i+2)^2}{9x_i}-\frac{(i-1)^2}{9x_{i-1}}\right)}\\ &= \frac{1}{x_1}+\sum_{i=2}^{n}{\frac{(i+2)^2}{9x_i}}-\sum_{i=1}^{n-1}{\frac{i^2}{9x_{i}}}\\ &= \frac{1}{x_1}+\frac{(n+2)^2}{9x_n}+\sum_{i=2}^{n-1}{\frac{4i+4}{9x_i}} \\ &\ge \frac49\left(\frac{2}{x_1}+\sum_{i=2}^{n-1}{\frac{i+1}{x_i}} + \frac{n+1}{x_n}\right) \end{align*}As desired. $~$ Now, we prove that $a\le \frac49$. Let $d_i=\tbinom{i+1}{2}$, then $x_i=\tbinom{i+2}{3}$. Then, we have \[2\sum_{i=1}^{\infty}{\frac{1}{i(i+1)}} \geq 3a \left( \sum_{i=1}^{\infty}{\frac{2}{i(i+2)}} \right)\]so $a\le \tfrac49$.