FAA2533 wrote: Let $a_1, a_2, \ldots, a_{2024}$ be a permutation of $1, 2, \ldots, 2024$. Find the minimum possible value of\[\sum_{i=1} ^{2023} \Big[(a_i+a_{i+1})\Big(\frac{1}{a_i}+\frac{1}{a_{i+1}}\Big)+\frac{1}{a_ia_{i+1}}\Big]\] Proposed by Md. Ashraful Islam Fahim BdMO=Bangladesh National Mathematical Olympiad

Is it the correct answer?

thegreatp.d wrote:

Is it the correct answer?

What's the configuration of your claim?

I think the answer 8092+2023/2024

thegreatp.d wrote:

Is it the correct answer?

This answer comes from the AM-GM minimum bound. But this can't be the answer since this minimum is only achieved when the numbers are all equal. Here they are not equal.

By AM-GM inequality, \begin{align*} (a_i+a_{i+1})\left(\frac{1}{a_i}+\frac{1}{a_{i+1}}\right)+\frac{1}{a_ia_{i+1}}&=4+\frac{(a_i-a_{i+1})^2+1}{a_ia_{i+1}}\\ \ge4+\frac{2|a_i-a_{i+1}|}{a_ia_{i+1}}&=4+2\left|\frac1{a_i}-\frac1{a_{i+1}}\right|. \end{align*}Let $\{a_p,a_q\}=\{1,2024\}$, where ${}p<q$, then we have \begin{align*} \sum_{i=1} ^{2023} \left[(a_i+a_{i+1})\left(\frac{1}{a_i}+\frac{1}{a_{i+1}}\right)+\frac{1}{a_ia_{i+1}}\right]&\ge8092+2\sum_{i=1}^{2023}\left|\frac1{a_i}-\frac1{a_{i+1}}\right|\\ \ge8092+2\sum_{i=p}^{q-1}\left|\frac1{a_i}-\frac1{a_{i+1}}\right|&\ge8092+2\left|\sum_{i=p}^{q-1}\left(\frac1{a_i}-\frac1{a_{i+1}}\right)\right|\\ =8092+2\left|1-\frac1{2024}\right|&=\frac{8191127}{1012}. \end{align*}Equality occurs when $a_i=i$ for all $i$.

Finally solved this one! But it's funny that the only solution (@above) has been posted this week and my solution is the same Here is the first and most important step in solving the problem, $$(x+y)\left(\frac 1x+\frac 1y\right)+\frac{1}{xy}=4+\frac{\left (1-|x-y|\right )^2}{xy}+2\left | \frac 1x-\frac 1y \right | \geq 4+2\left | \frac 1x-\frac 1y \right |.$$ Let $S$ denote the sum we wish to minimize. We have, $$S\geq \sum_{i=1}^{2023}\left (4+2 \left |\frac{1}{a_i}-\frac{1}{a_{i+1}}\right | \right)=8092+2 \sum_{i=1}^{2023} \left |\frac{1}{a_i}-\frac{1}{a_{i+1}}\right |.$$Now we shall concentrate on minimizing $\sum_{i=1}^{2023} \left |\frac{1}{a_i}-\frac{1}{a_{i+1}}\right |$. Suppose that $a_k=1$ for some $k \notin \{1,2024\}$. Note that \begin{align*} \sum_{i=1}^{2023} \left |\frac{1}{a_i}-\frac{1}{a_{i+1}}\right | & \geq \left |\frac{1}{a_2}-\frac{1}{a_1}+\frac{1}{a_3}-\frac{1}{a_2}+\dots+\frac{1}{a_k}-\frac{1}{a_{k-1}}\right |+\left |\frac{1}{a_k}-\frac{1}{a_{k+1}}+\dots+\frac{1}{a_{2023}}-\frac{1}{a_{2024}} \right | \\ &=1-\frac{1}{a_1}+1-\frac{1}{a_{2024}} \geq 2-\frac 12-\frac 13 > 1-\frac{1}{2024}.\end{align*} Suppose $a_1=1$ (after re-indexing, $a_{2024}=1$ is the same as this). Let $a_n=2024$ for some $n\leq 2024$. Note that $$\sum_{i=1}^{2023} \left |\frac{1}{a_i}-\frac{1}{a_{i+1}}\right | \geq \left | \sum_{i=1}^{n-1} \left ( \frac{1}{a_i}-\frac{1}{a_{i+1}}\right ) \right |+\sum_{i=n}^{2023} \left |\frac{1}{a_i}-\frac{1}{a_{i+1}}\right | \geq \left |\frac{1}{a_1}-\frac{1}{a_n}\right |=1-\frac{1}{2024}.$$ Back to our problem, we now have $S\geq 8092+2\left (1-\frac{1}{2024} \right)=8094-\frac{1}{1012}$. We can check that equality holds when $a_i=i$ for every $i$. Hence the desired minimum is $\boxed{8094-\frac{1}{1012}}$.