Determine a constant that satisifies an inequality Determine the minimum real number $c$ such that $$\sum_ {k=1}^n \frac {a_k}{c^{a_1+a_2+ \dots +a_k}} \le 1,$$ for every positive integer $n$ and positive real numbers $a_1,a_2, \dots, a_n$.
 A: In order to get some insight into the problem we substitute $\gamma=\dfrac{1}{c}$ and restrict our attention to positive integers $a_k\gt0$ and a fixed $n\in\mathbb N$. We compute $b_k:=\sum_{j=1}^ka_j\gt0$. Then the inequality can be written to
$$\tag{1}\sum_{k=1}^na_k\gamma^{b_k}\le1.$$
All the numbers $b_k\gt0$ are integers and the left side is a polynomial. Because


*

*it is $0$ for $\gamma=0$ 

*the left side grows monotonically with growing $\gamma$ since all $a_k\gt0$

*it is at least $1$ for $\gamma=a_1^{-1/a_1}$ since $\sum_{k=1}^na_k\gamma^{b_k}\ge a_1\cdot (a_1^{-1/a_1})^{b_1}= a_1\cdot a_1^{-b_1/a_1}=a_1\cdot a_1^{-a_1/a_1}=1$


there exists exactly $1$ solution $\gamma_0$. This solution $\gamma_0$ is a zero of a polynomial if we write the inequality $(1)$ as an equation. There exists a closed solution to the problem if the Galois group of the polynomial is solvable. Then this solution can be expressed as a sum of nested sums of radicals (or roots). However at best it is hard to determine the Galois group. And if the Galois group is not solvable there exists no known closed expression of a solution. If we release the restriction that the positive numbers $a_k$ are integers we get from bad to worse and therefore we cannot expect a closed formula for the solution $\gamma_0$.
The explanations above have already indicated how to find the solution $\gamma_0$. The three statements above also hold if only $a_k\gt0$. Therefore we start with $\gamma_{l,0}=0$ and $\gamma_{u,0}=a_1^{-1/a_1}$. Then we take the value $\gamma_{m,k}:=\dfrac{\gamma_{l,k}+\gamma_{u,k}}{2}$ for $k=0$ and compute the left side of the inequality $(1)$ for $\gamma_{m,k}$. If the inequality holds we take $\gamma_{l,k+1}=\gamma_{m,k}$ and $\gamma_{u,k+1}=\gamma_{u,k}$ in the next step. Otherwise we take $\gamma_{l,k+1}=\gamma_{l,k}$ and $\gamma_{u,k+1}=\gamma_{m,k}$ and consecutively rerun the process with $k+1$ until we get a sufficiently small error. We close in on the solution with an error less than $\dfrac{\gamma_{u,0}}{2^k}$ and this is exponentially fast.
Now we take the numbers $a_1,\dots,a_n$. For any $0\lt n_1\lt n_2\le n$ it is apparent that
$$\sum_{k=1}^{n_1}a_k\gamma^{b_k}\le\sum_{k=1}^{n_2}a_k\gamma^{b_k}.$$
Therefore if the inequality $(1)$ holds for $n$ it holds for all $\hat n\lt n$. If $n$ is inifinite we have the numbers $a_1,a_2,a_3,\dots$ and we can take the same procedure above with $n=\infty$. However we must additionally determine the $\gamma\ge0$ for which the series 
$$\sum_{k=1}^{\infty}a_k\gamma^{b_k}$$
converges. Moreover we cannot sum up infinitely many summands and must determine the error when we restrict the sum to a finite number of summands. This can be really hard for a Taylor series. The problem in your question is even harder and therefore a general solution cannot be given in the case $n=\infty$.
A: Let $x>1$ and $f(x)$ be the smallest real number $y$ satisfying that, for any integer $n$ and $n$ numbers $a_1, a_2, \cdots, a_n$, $\sum_ {k=1}^n \frac {a_k}{x^{a_1+a_2+ \dots +a_k}} \le y$. Let's find a closed form expression for $f(x)$.
First of all, $f(x) \ge \lim_{\epsilon \rightarrow 0}\frac{\epsilon}{x^\epsilon-1}=\frac{1}{\ln x}$ because the value of infinite sum $\sum_{k=1}^{\infty}\frac{\epsilon}{x^{k\epsilon}}$ is $\frac{\epsilon}{x^\epsilon-1}$ and this is strictly decreasing function of $\epsilon$ when $x>1$. We want to prove that $f(x)=\frac{1}{\ln x}$.
For any $x>1$, let's suppose that there are integer $M$ and $a_1, a_2, \cdots, a_M$ satisfying $\sum_ {k=1}^M \frac {a_k}{x^{a_1+a_2+ \dots +a_k}}>\frac{1}{\ln x}$. Let's also suppose that among these sequences, $a_1, a_2, \cdots, a_m$ is the shortest.
We know that $m>1$ because $\frac{a_1}{x^{a_1}}\le\frac{1}{e\ln x}$. By the assumption of shortest sequence, we also know that the sum generated by the sequence $a_2, a_3, \cdots, a_m$ does not exceed $\frac{1}{\ln x}$, in other words, $\sum_ {k=2}^m \frac {a_k}{x^{a_1+a_2+ \dots +a_k}}\le \frac{1}{\ln x}$. Let $A=\sum_ {k=2}^m \frac {a_k}{x^{a_1+a_2+ \dots +a_k}}$. By our assumption, $A<\frac{1}{\ln x}<\frac{A+a_1}{x^{a_1}}$ and $a_1$ maximizing $\frac{A+a_1}{x^{a_1}}$ is $a_1=\frac{1}{\ln x}-A$. Now we get$$\frac{1}{\ln x}<\frac{A+a_1}{x^{a_1}}\le\frac{x^A}{e\ln x}\le\frac{x^{1/\ln x}}{e\ln x}=\frac{1}{\ln x}$$which is contradiction. Therefore, our assumption of existence of $M$ and sequence $a_1, a_2, \cdots, a_M$ satisfying $\sum_ {k=1}^M \frac {a_k}{x^{a_1+a_2+ \dots +a_k}}>\frac{1}{\ln x}$ is wrong and it is true that $f(x)=\frac{1}{\ln x}$.
We are looking for smallest $x$ satisfying $f(x) \le 1$. The answer should be $e$.
