What’s interesting, a problem in inequality marathon: aops.com/community/c6h3169191p28848444

This might be the hardest problem in 2024 CTST, I heard only 1 person write something, and it is the only problem Chunji Wang failed in TST Round 2. I've heard this have something to do with this famous problem, the weaker bound of this is found in PFTB: Let integer $n\ge 3,$ positive real numbers $x_1,x_2,\ldots ,x_n$ satsify $$(x_1+x_2+\cdots +x_n)\left(\frac 1{x_1}+\frac 1{x_2}+\cdots +\frac 1{x_n}\right)=n^2+1.$$Prove that $$(x_1^2+x_2^2+\cdots +x_n^2)\left(\frac 1{x_1^2}+\frac 1{x_2^2}+\cdots +\frac 1{x_n^2}\right)\ge n^2+4+\frac 4{n^2}.$$ I know how to prove the above problem, by this and Lagrange equation we can get $$\sum_{1\le i<j\le n}\left(\sqrt{\frac{x_i}{x_j}}-\sqrt{\frac{x_j}{x_i}}\right)^2=1.$$Also \begin{align*}4+\frac 4{n^2}&\le\sum_{1\le i<j\le n}\left({\frac{x_i}{x_j}}-{\frac{x_j}{x_i}}\right)^2\\&=\sum_{1\le i<j\le n}\left(\sqrt{\frac{x_i}{x_j}}-\sqrt{\frac{x_j}{x_i}}\right)^2\left[\left(\sqrt{\frac{x_i}{x_j}}-\sqrt{\frac{x_j}{x_i}}\right)^2+4\right]\\&=4+\sum_{1\le i<j\le n}\left(\sqrt{\frac{x_i}{x_j}}-\sqrt{\frac{x_j}{x_i}}\right)^4.\end{align*}Therefore $$\frac{n^2}4\sum_{1\le i<j\le n}\left(\sqrt{\frac{x_i}{x_j}}-\sqrt{\frac{x_j}{x_i}}\right)^4\ge\left[\sum_{1\le i<j\le n}\left(\sqrt{\frac{x_i}{x_j}}-\sqrt{\frac{x_j}{x_i}}\right)^2\right]^2$$Seems like if $a_{i,j}=\sqrt{\frac{x_i}{x_j}}-\sqrt{\frac{x_j}{x_i}}$ there is some relationship... when $n$ is even it can be proved this way.

Hmmm well it's hard because it relies on some kind of magic trick. The kind of problem I don't enjoy personnaly.

Related issues: A regular 2n+1-sided polygon and all its diagonals can be divided into several non intersecting union sets of convex polygons

This is a translation of a solution by Cheng Jiang (Perfect IMO Score) sent to me by @EthanWYX2009. Thanks for @YaoAOPS for helping me translate it. We initially will prove the case where $n$ is odd, the case where $n$ is even can be obtained in a rather straightforward manner after that, e.g, ignore one index and sum everything. Let $n = 2k + 1$. We begin by two lemmas Lemma. Let $M$, $x_1$, $x_2$, $\dots$, $x_n$ be non-negative real numbers with $M \ge x_1 + x_2 + \cdots + x_n$, then \[2(M^4 + \sum_{i=1}^{n}x_i^{4}) \ge (M^2 + \sum_{i=1}^{n}x_i^2)^2\]Proof. Notice this is equivalent to showing $(M^2 - \sum_{i=1}^{n}x_i^{2})^{2} \ge 4\sum_{i<j}x_i^{2}x_j^{2}$ which is clear. Lemma. A set of ordered pairs $X$ is called a cycle when there is a sequence of indexes $i_1$, $i_2$, $\dots$, $i_n$ such that $X = \{(i_{k}, i_{k+1}) | 1 \le k \le n - 1) \cup \{(i_1, i_{n})\}$. Let $n = 2k+1$, we have that $[n]$ can be decomposed into $\frac{k(k+1)}{2}$ disjoint cycles. Proof. This is longer to state than it is to prove. Use induction by creating $k$ cycles with elements $1$, $k+1$, then apply induction hypothesis. Now, as in the above lemma, consider the decomposition of the indexes $[n]$ in cycles to be $C_1$, $C_2$, $\dots$, $C_{\frac{k(k+1)}{2}}$ and it is easy to verify that on each cycle there's an element that is at least the size of the sum of the other elements, we have \begin{align*} k(k+1)\sum_{1\le i<j\le n}a_{i,j}^4 &= \\ \frac{k(k+1)}{2}\sum_{i=1}^{\frac{k(k+1)}{2}}\sum_{(i,j) \in C_i}2a_{i, j}^{4} &\ge^{lemma}\\ \frac{k(k+1)}{2}\sum_{i=1}^{\frac{k(k+1)}{2}}\left(\sum_{(i,j) \in C_i}a_{i,j}^{2}\right)^{2} &\ge^{CS}\\ \left(\sum_{i=1}^{\frac{k(k+1)}{2}}\sum_{(i,j) \in C_i}a_{i,j}^{2}\right)^{2} &= \left(\sum_{1 \le i < j \le n}a_{i,j}^{2}\right)^{2}\\ \end{align*}Concluding the proof $\blacksquare$.

Wild solution created by proposer Jingjun Han: We use induction to prove a stronger result(Original version is $P(n,1)$): Quote: Let integer $n,l\ge 2,$ $\tbinom n2$ nonnegative real numbers $a_{i,j}$ satisfy $ a_{i,j}+a_{j,k}\le a_{i,k}$ holds for all $1\le i <j<k\le n$. Proof $$P(n,l):=\left\lfloor\frac{(n+l-1)^2}4\right\rfloor\left(\sum_{1\le i<j< n}a_{i,j}^4+l\sum_{i=1}^{n-1}a_{i,n}^4\right)-\left(\sum_{1\le i<j< n}a_{i,j}^2+l\sum_{i=1}^{n-1}a_{i,n}^2\right)^2\ge 0.$$ Base case $n=2$ is trivial. Consider $n\ge 3.$ Define $$f(a_1,\ldots ,a_{n-1}):=\left\lfloor\frac{(n+l-1)^2}4\right\rfloor\left(\sum_{1\le i<j< n}a_{i,j}^4+l\sum_{i=1}^{n-1}a_{i}^4\right)-\left(\sum_{1\le i<j< n}a_{i,j}^2+l\sum_{i=1}^{n-1}a_{i}^2\right)^2.$$If $a_{i,j}\le a_i-a_j$ holds when $1\le i<j<n,$ then $\sum_{1\le i<j< n}a_{i,j}^2\le \sum_{1\le i<j< n}(a_i-a_j)^2=(n-1)\sum_{i=1}^{n-1}a_{i}^2-\left(\sum_{i=1}^{n-1}a_{i}\right)^2.$ Let $G(y):=f(a_{1,n}+y,\ldots ,a_{n-1,n}+y).$ We have \begin{align*}G'(y)&=\sum_{i=1}^{n-1}\frac{\partial f}{\partial x_i} \\&=4\left\lfloor\frac{(n+l-1)^2}4\right\rfloor l\sum_{i=1}^{n-1}a_{i}^3-4l\left(\sum_{1\le i<j< n}a_{i,j}^2+l\sum_{i=1}^{n-1}a_{i}^2\right)\sum_{i=1}^{n-1}a_{i}

, we can adjust $(a_1,\ldots ,a_{n-1})$ to $(u,u,\ldots ,u,0,\ldots ,0).$ Plug in we can easily get $G'(y)\ge 0$ when $y+a_{n-1,n}\ge 0.$ Therefore $G(0)\ge G(-a_{n-1,n}),$ so WLOG $a_{n-1,n}=0.$ Now let $A_p:=\sum_{1\le i<j< n-1}a_{i,j}^p,$ $B_p:=\sum_{i=1}^{n-2}a_{i,n-1}^2,$ $C_p:=\sum_{i=1}^{n-2}a_{i,n}^2$ for $p=2,4.$ We need to prove $$\left\lfloor\frac{(n+l-1)^2}4\right\rfloor (A_4+B_4+lC_4)\ge (A_2+B_2+lC_2)^2.$$By induction hypothesis $P(n-1,l+1),$ $$\left\lfloor\frac{(n+l-1)^2}4\right\rfloor (A_4+(l+1)B_4)\ge (A_2+(l+1)B_2)^2,$$$$\left\lfloor\frac{(n+l-1)^2}4\right\rfloor (A_4+(l+1)C_4)\ge (A_2+(l+1)C_2)^2.$$Therefore by Cauchy Schwartz Inequality \begin{align*}(A_2+B_2+lC_2)^2&=\frac 1{(l+1)^2}\left((A_2+(l+1)B_2)+l(A_2+(l+1)C_2)\right)^2\\&\le \frac 1{l+1}\left((A_2+(l+1)B_2)^2+l(A_2+(l+1)C_2)^2\right)\\&\le\frac 1{l+1}\left\lfloor\frac{(n+l-1)^2}4\right\rfloor\left((A_4+(l+1)B_4)+l(A_4+(l+1)C_4)\right)\\&=\left\lfloor\frac{(n+l-1)^2}4\right\rfloor (A_4+B_4+lC_4).\Box\end{align*}