Estonian Mathematical Olympiad

Multiplying out $(k+1{)}^{10}$ yields among others the terms $k^{10}$ and $10k^9$. All of the other terms are positive for a positive $k$. Thus for all positive integers $k$ we have $(k+1{)}^{10}>k^{10}+10k^9$. Combining this for $k=2022,2021,\dots ,2,1$ yields $$\begin{array}{rl}2023^{10}& >2022^{10}+10\cdot 2022^9\\ & >2021^{10}+10\cdot 2021^9+10\cdot 2022^9\\ & >\dots \\ & >1^{10}+10\cdot 1^9+10\cdot 2^9+\cdots +10\cdot 2021^9+10\cdot 2022^9\\ & >10\cdot (1^9+2^9+3^9+\cdots +2022^9).\end{array}$$

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Alternative solution.

We will prove that for all integers $a,n$ greater than 1, $$a^n>n\cdot (1^{n-1}+2^{n-1}+3^{n-1}+\cdots +(a-1{)}^{n-1});(4)$$ where taking $a=2023$, $n=10$ yields the desired result. We will prove this via mathematical induction on the variable $a$. For $a=2$, the inequality $2^n>n\cdot 1^{n-1}=n$ is well known. For the induction step we will show that if (4) holds, then so does $$(a+1{)}^n>n\cdot (1^{n-1}+2^{n-1}+3^{n-1}+\cdots +(a-1{)}^{n-1}+a^{n-1}).$$ We will show that the left hand side increases by at least as much as the right hand side or $(a+1{)}^n-a^n\ge n\cdot a^{n-1}$. Using the formula for the difference of powers together with $a+1>a$ yields $$\begin{array}{rl}(a+1{)}^n-a^n& =((a+1)-a)((a+1{)}^{n-1}+(a+1{)}^{n-2}a+\cdots +(a+1)a^{n-2}+a^{n-1})\\ & >1\cdot (a^{n-1}+a^{n-1}+\cdots +a^{n-1}+a^{n-1})=n\cdot a^{n-1},\end{array}$$ which proves the induction step and thus finishes the proof.

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Alternative solution.

As the function $y=x^9$ is increasing in $[1;2023]$, we have $$1^9+2^9+3^9+\cdots +2022^9<{\int }_1^{2023}{x}^{9}dx.$$ As the integral of $y=x^9$ is $y=\frac{1}{10}{x}^{10}$, the Newton-Leibniz formula gives $${\int }_1^{2023}{x}^{9}dx=\frac{1}{10}\cdot 2023^{10}-\frac{1}{10}\cdot 1^{10}<\frac{1}{10}\cdot 2023^{10}.$$ Altogether, $10\cdot (1^9+2^9+3^9+\cdots +2022^9)<2023^{10}$ as desired.