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 $a_1,a_2, \dots, a_n$.

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 $e$.
