Let $n$ be a positive integer, and $a_1, a_2, \dots, a_{2n}$ be a sequence of positive real numbers whose product is equal to $2$. For $k = 1, 2, \dots, 2n$, set $a_{2n + k} = a_k$, and define $$ A_k = \frac{1 + a_k + a_k a_{k + 1} + \dots + a_k a_{k + 1} \cdots a_{k + n - 2}}{1 + a_k + a_k a_{k + 1} + \dots + a_k a_{k + 1} \cdots a_{k + 2n - 2}}. $$ Suppose that $A_1, A_2, \dots, A_{2n}$ are pairwise distinct; show that exactly half of them are less than $\sqrt{2} - 1$.
Problem
Source: 2022 Pan-African Mathematics Olympiad Problem 3
Tags: algebra, inequalities
China_BW
10.07.2022 13:37
done Ai and Ai+n has a relationship
Rukevwe
11.07.2022 22:31
For $k=1, 2, \cdots , 2n,$ let $b_k=1+a_k+a_ka_{k+1}+\cdots+a_ka_{k+1}\cdots a_{k+n-2}$ and let $c_k=a_ka_{k+1}\cdots a_{k+n-1}$.
Then for $k=1, 2, \cdots , n,$ we have $c_kc_{k+n}=a_ka_{k+1}\cdots a_{k+2n-1}=2$ and $b_k+c_kb_{k+n}=1+a_k+a_ka_{k+1}+\cdots+a_ka_{k+1}\cdots a_{k+2n-2}$, so $A_k=\frac{b_k}{b_k+c_kb_{k+n}}$. Similarly $A_{k+n}=\frac{b_{k+n}}{b_{k+n}+c_{k+n}b_k}$. Then,
\begin{align*}
(1+A_{k})(1+A_{k+n})&=\left[\frac{c_{k}b_{k+n}+2b_{k}}{b_{k}+c_{k}b_{k+n}}\right] \left[\frac{c_{k+n}b_{k}+2b_{k+n}}{b_{k+n}+c_{k+n}b_{k}}\right] \\
&=\left[\frac{c_{k}b_{k+n}+c_kc_{k+n}b_{k}}{b_{k}+c_{k}b_{k+n}}\right] \left[\frac{c_{k+n}b_{k}+c_kc_{k+n}b_{k}}{b_{k+n}+c_{k+n}b_{k}}\right] \\
&=\left[\frac{c_{k}(b_{k+n}+c_{k+n}b_{k})}{b_{k}+c_{k}b_{k+n}}\right] \left[\frac{c_{k+n}(b_{k}+c_{k}b_{k+n})}{b_{k+n}+c_{k+n}b_{k}}\right] \\
&=c_kc_{k+n}=2
\end{align*}Hence, since $A_k \neq A_{k+n},$ we have that either $1+A_k<\sqrt{2}, 1+A_{k+n}>\sqrt{2}$ or $1+A_k>\sqrt{2}, +A_{k+n}<\sqrt{2}$. That is, exactly one of $A_k, A_{k+n}$ is less than $\sqrt{2} - 1$. Doing this for $k=1, 2, \cdots , n$ gives the desired result.
ZNatox
08.01.2023 02:12
We have: $A_k=\frac{1+a_k+a_ka_{k+1}+\cdots +a_ka_{k+1}\cdots a_{k+n-2}}{1+a_k+a_ka_{k+1}+\cdots +a_ka_{k+1}\cdots a_{k+2n-2}}=\frac{1}{1+\frac{a_ka_{k+1}\cdots a_{k+n-1}+\cdots +a_ka_{k+1}\cdots a_{k+2n-2}}{1+a_k+a_ka_{k+1}+\cdots +a_ka_{k+1}\cdots a_{k+n-2}}}.$ Which gives us: $\frac1{A_k}-1=\sqrt2×\frac{f(n+k)}{f(k)}$, where $f(i)=\frac{1+a_i+a_ia_{i+1}+\cdots +a_ia_{i+1}\cdots a_{i+n-2}}{\sqrt{a_ia_{i+1}\cdots a_{i+n-1}}}$ (We used the condition on the product of the $a_is$ to get this equality). Finally this gives us that $(\frac1{A_k}-1)(\frac1{A_{k+n}}-1)=2$, and since the $A_is$ are distinct we can conclude that for each $i$: one of $\frac1{A_i}-1$ and $\frac1{A_{i+n}}-1$ is smaller than $\sqrt2$ while the other one is greater than it, which gives us what we wanted. Done .