Tolin'go A.K.

We took several positive numbers and constructed a sequence from them as follows: $a_{1}$ is the sum of the original numbers, $a_{2}$ is the sum of the squares of the original numbers, $a_{3}$ is the sum of the cubes of the original numbers, and so on.

a) Could it happen that the sequence decreases up to $a_{5}$ ($a_{1}>a_{2}>a_{3}>a_{4}>a_{5}$), and starting from $a_{5}$ it increases ($a_{5}<a_{6}<a_{7}<\ldots$)?

b) And could it happen the other way around: the sequence increases up to $a_{5}$, and starting from $a_{5}$ it decreases?
a) Example 1. Take the number 2 and 1024 numbers, each equal to $1 / 2$. Then $a_{n}=2^{n}+1024 \cdot 2^{-n}=32\left(2^{n-5}+2^{5-n}\right)$. The sum of two positive reciprocal numbers is the smallest when they are closest to each other. Therefore, the constructed sequence decreases up to $n=5$, and then increases.

Example 2. Take $a_{n}=1.02^{n}+0.5^{n}$. Note that $a_{n+1}>a_{n} \Leftrightarrow 0.02 \cdot 1.02^{n}>0.5^{n+1} \Leftrightarrow 2.04^{n}>25 \Leftrightarrow n \geq 5$. That is, the sequence decreases up to $n=5$, and then increases.

b) Suppose this happened.

First method. Then $a_{n}1$, otherwise the sequence would never increase. Note that $a_{n}>x^{n}$ for any $n$. And the sequence $x^{n}$ is unbounded. Contradiction.

Second method. Then $a_{4}<a_{5}$ and $a_{6}<a_{5}$, so $a_{4}+a_{6}<2 a_{5}$. This means that $x^{4}+x^{6}<2 x^{5}$ for at least one of the original numbers $x$. But this contradicts the Cauchy inequality.

## Answer

a) It could; b) it could not.