KoMaL A.394
Let $a_1,a_2,\dots,a_N$ be nonnegative reals, not all 0. Prove that there exists a sequence $1=n_0<n_1<\dots<n_k=N+1$ of integers such that\[n_1a_{n_0}+n_2a_{n_1}+\dots+n_ka_{n_{k-1}}<3(a_1+a_2+\dots+a_N).\]
Solution.Define $a_{N+2}+a_{N+3}=\dots=0$ as well and look for an infinite sequence $1=n_0<n_1<\dots$ of integers for which
\[\sum_{i=1}^\infty n_ia_{n_{i-1}} < 3\sum_{n=1}^\infty a_n.\]
Define the function $f:[1,\infty)\to R$ such that $f(x)=a_{[x]}$.
The required sequence is constructed randomly. Take a random variable $t\in[0,1]$ of uniform distribution. Set $n_0=1$ and $n_i=\left[2\cdot e^{i-1+t}\right]$ for $i=1,2,\dots$. Then\begin{aligned}
&E\left(\sum_{i=1}^\infty n_ia_{n_{i-1}}\right) \le E(n_1)\cdot a_0 + E \left(\sum_{i=2}^\infty2\cdot e^{i-1+t}f(2\cdot e^{i-2+t})\right)\\
=&E(n_1)\cdot a_0 + \sum_{i=2}^\infty\int_0^1 \big(2\cdot e^{i-1+t}f(2\cdot e^{i-2+t})\big)dt\\
=&E(n_1)\cdot a_0 + \int_0^\infty 2\cdot e^{u+1}f(2\cdot e^u)du\\=&E(n_1)\cdot a_0 + e\int_2^\infty f(x)dx\\
=&E(n_1)\cdot a_0 + e\sum_{n=2}^Na_n.
\end{aligned}The value of $n_1$ is 2,3,4 or 5 if $t<\ln\frac32$, $\ln\frac32\le t<\ln\frac42$, $\ln\frac42\le t<\ln\frac52$, or $\ln\frac52\le t<1$, respectively. Therefore
\[E(n_1)= \left(\ln\frac32\right)\cdot2 + \left(\ln\frac42-\ln\frac32\right)\cdot3 + \left(\ln\frac52-\ln\frac42\right)\cdot4 + \left(1-\ln\frac52\right)\cdot5 \approx 2.985<3\]
and
\[E\left(\sum_{i=1}^\infty n_ia_{n_{i-1}}\right) < 3a_1 + e(a_2+\dots+a_N).\]
There always exists a sequence $1=n_0<n_1<...<n_k=N+1$ such that
\[n_1a_{n_0}+n_2a_{n_1}+\dots+n_ka_{n_{k-1}}<3a_1+e(a_2+a_3+...+a_N).\] math.stackexchange.com/questions/1060803 |