Baltic Way 2023 Shortlist

Answer: $\frac{17}{2}$.

Note that $2023=7\cdot 17^2=7\cdot 289=17\cdot 119$. Using AGM on the numbers $x^{2023},1,1,\dots ,1$, where there are 288 ones, yields $$x^{2023}+288\ge 289\sqrt[289]{x^{2023}}=289x^7.$$ Analogously, using AGM on the numbers $x^{2023},1,1,\dots ,1$, where there are 118 ones, yields $$x^{2023}+118\ge 119\sqrt[119]{x^{2023}}=119x^{17}.$$ In both inequalities the equality holds if and only if $x=1$. Adding them yields $2x^{2023}+406\ge 289x^7+119x^{17}$, which yields $2(x^{2023}+203)\ge 17(17x^7+7x^{17})$. Therefore $\frac{x^{2023}+203}{17x^7+7x^{17}}\ge \frac{17}{2}$, where $x=1$ is still the equality case. Therefore the smallest possible value of $\frac{x^{2023}+203}{17x^7+7x^{17}}$ is $\frac{17}{2}$.

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Alternative solution.

Denote $$f(x)=\frac{x^{2023}+203}{17x^7+7x^{17}}.$$ Then $$f^{\prime }(x)=119x^6\cdot \frac{118x^{2023}+288x^{2023}-203x^{10}-203}{(17x^7+7x^{17}{)}^2}.$$ To show that $f$ obtains its minimal value at $x=1$, we will show that $f^{\prime }$ is negative for $0<x<1$, zero at $x=1$ and positive for $x>1$. This is equivalent to showing the same for $118x^{2023}+288x^{2023}-203x^{10}-203$. We see that its value is indeed zero at $x=1$ and that (e.g. by computing the geometric sums on the right-hand side) $$\begin{array}{rl}& 118x^{2023}+288x^{2023}-203x^{10}-203\\ & =(x-1)\cdot (118(x^{2022}+\cdots +x^{2023}))\\ & +406(x^{2022}+\cdots +x^{10})+203(x^9+\cdots +x+1).\end{array}$$ The second factor is clearly positive for all positive $x$. So $f^{\prime }$ is negative for $0<x<1$, zero at $x=1$ and positive for $x>1$, as desired. Thus the minimal value of $f(x)$ is $$f(1)=\frac{1+203}{17\cdot 1+7\cdot 1}=\frac{204}{24}=\frac{17}{2}.$$

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Alternative solution.

Like in Solution 2, reduce the problem to having to show that $118x^{2033}+288x^{2023}-203x^{10}-203$ is negative for $0<x<1$ and positive for $x>1$. For $0<x<1$, we have $x^{2033}<x^{2023}<x^{10}<1$, so $118x^{2033}+85x^{2023}-203x^{10}<0$ and $203x^{2023}-203<0$. Adding those yields the desired result. For $x>1$, we have $x^{2033}>x^{2023}>x^{10}>1$, so $118x^{2033}+85x^{2023}-203x^{10}>0$ and $203x^{2023}-203>0$. Adding those yields the desired result.