Can you correct LaTeX?

WolfusA wrote: Can you correct LaTeX? I edited

For each $i\in [n]$, let $\ell_i=\sum_{j=1}^{i}{\frac{1}{a_{i,j}}} +\sum_{j=i}^{n}{a_{j,i}}-a_{i,i}$ and note that $a_{i,i}=1$. We want to maximize $S=\sum_{k=1}^{n}{\frac{1}{\ell_k}}$ where $a_{i,j}\in \mathbb{R}^+$ for all $i,j\in [n]$ with $i\geqslant j$. Let $b_{i,j}$ where $i,j\in [n]$ with $i\geqslant j$ be the tuples of $\frac{n(n+1)}{2}$ values that maximize $S$ (we'll also show that the maximum attain when no terms is $0$ or $\infty$). For each $i,j\in [n]$ with $i>j$, the terms in $S$ that are not independent of variable $b_{i,j}$ are $\frac{1}{\ell_i}+\frac{1}{\ell_j}$. Let $c=\sum_{k=1}^{j}{\frac{1}{b_{k,j}}} +\sum_{k=j}^{n}{b_{k,j}}-b_{j,j}-b_{i,j}=\ell_j -b_{i,j}$ and $d=\sum_{k=1}^{i}{\frac{1}{b_{k,i}}} +\sum_{k=i}^{n}{b_{k,j}}-b_{i,i}-\frac{1}{b_{i,j}}=\ell_i -\frac{1}{b_{i,j}}$. Note that $c> b_{j,j} =1$ and $d> b_{i,i}=1$. We've $$\frac{1}{\ell_i} +\frac{1}{\ell_j} =\frac{1}{d+\frac{1}{b_{i,j}}} +\frac{1}{c+b_{i,j}} =\frac{b_{i,j}}{db_{i,j}+1} +\frac{1}{c+b_{i,j}}.$$So, $b_{i,j}$ is the value of $x\in \mathbb{R}^+$ that maximize $f(x)=\frac{x}{dx+1}+\frac{1}{c+x}$. We've $f'(x)= \frac{1}{(dx+1)^2} -\frac{1}{(c+x)^2}$, so $f'(x)=0\implies x\in \{ 0,\frac{c-1}{d-1},\infty\}$. Since $f(\frac{c-1}{d-1})= \frac{c+d-2}{cd-1}$ is greater than both $f(0)=\frac{1}{c}$ and $f(\infty )=\frac{1}{d}$. So, $b_{i,j}=\frac{c-1}{d-1}\implies b_{i,j}=\frac{\ell_j -b_{i,j}-1}{\ell_i -\frac{1}{b_{i,j}} -1}\implies b_{i,j}=\frac{\ell_j}{\ell_i}$. Consider $\ell_1$, we have $\ell_1 =1+\sum_{k=2}^{n}{b_{k,1}} =1+\sum_{k=2}^{n}{\frac{\ell_1}{\ell_k}}\implies 1=\sum_{k=1}^{n}{\frac{1}{\ell_k}}$. Hence, $1\geqslant S$ for any tuple of $\frac{n(n+1)}{2}$ positive real numbers. Equality holds when all terms equal to $1$.

ThE-dArK-lOrD wrote: ...Let $b_{i,j}$ where $i,j\in [n]$ with $i\geq j$ be the tuples of $\frac{n(n+1)}{2}$ values that maximize $S$ ... Hence, $b_{i,j}=\frac{\ell_j -b_{i,j}-1}{\ell_i -\frac{1}{b_{i,j}} -1}\implies b_{i,j}=\frac{\ell_j}{\ell_i}$. Consider $\ell_1$, we have $\ell_1 =1+\sum_{k=2}^{n}{b_{k,1}} =1+\sum_{k=2}^{n}{\frac{\ell_1}{\ell_k}}\implies 1=\sum_{k=1}^{n}{\frac{1}{\ell_k}}$. Hence, $1\geq S$ for any tuple of $\frac{n(n+1)}{2}$ positive real numbers. Equality holds at all terms equal to $1$. There is a flaw here. First, you cannot choose values $b_{ij}$ that maximize $S$, since they may not exist. Trying another approach on the same idea also fails. Well, if you fix all variables, except say $b_{ij}$ , indeed the max value attains when it holds $b_{i,j}=\frac{\ell_j}{\ell_i}$. But suppose you fixi for example $b_{21}$ satisfying $b_{2,1}=\frac{\ell_1}{\ell_2}$. Then when the next one $b_{31}$ is fixed to satisfy $b_{3,1}=\frac{\ell_1}{\ell_3}$, we'll break the first one. I mean, the way it is written, it essentially uses that there exist values $b_ij$ that maximize $S$, which you couldn't know a priori, because the domain is not compact. EDIT: I think, it can be fixed, but it needs essential effort to justify the idea.

Induction on $n$ combined with the approach as in #4 post (ThE-dArK-lOrD). Let $A$ be the matrix $(a_{ij}), i,j\in [n]$ and $S(A) := \sum_{i=1}^{n}\dfrac {1}{c_i}$. Suppose we have proven $S(A)\leq 1$ for all complying matrices with order less than $n\times n$ and we want to prove it for any $n\times n$ matrix $A$, which satisfies the conditions. Assume for the sake of contradiction, it's not true, that's there exists a matrix $A_0$ for which $S(A_0)=1+\varepsilon\,,\,\varepsilon>0$. Consider the set $M$ of matrices $A$ ,complying the problem's statement with $S(A)\geq 1+\varepsilon/2$. Let's prove, there exist $r,R\in \mathbb{R}^+$, such that $a_{ij}\in [r,R]$ for any matrix $A\in M$. Indeed, assume for example $a_{k\ell}$ is a small one, then $a_{\ell k}$ is a big one, hence $\frac{1}{c_k}$ is very small. Thus if we delete the $k$-th row and $k$-th column of $A$ we will get a $n-1\times n-1$ matrix $A'$ also complying the conditions for which $S(A')\leq 1$. Since $S(A)\leq S(A')+\frac{1}{c_k}$ we will arrive at a contradiction with $S(A)\geq 1+\varepsilon/2$ if we choose small enough $r$ and big enough $R$. Thus, we proved that for those $A$ for which $S(A)\geq 1+\varepsilon/2$ , the corresponding elements $a_{ij}$ are bounded in some interval $[r,R]$. Since $S$ is continuous, it attains its maximum on a compact set, thus there exist a matrix $(b_{ij})$ as in the post #4, that maximizes $S$. Following the same approach, those $b_{ij}$ must satisfy $b_{ij}=\frac{c_j}{c_i}$ , hence: $$c_1 =c_1\cdot \sum_{k=1}^n \frac{1}{c_k}=c_1 S((b_{ij})) $$It means $S((b_{ij}))=1$, which is a contradiction with the assumption that $S$ can get values bigger than $1$.

Assume that true for $n$ then prove for $n+1$. Let $d_i=a_{i,n+1}$ we can easily get ,$\sum_{i=1}^{n+1}\dfrac {1}{c_i}=\sum_{i=1}^{n}\dfrac {1}{c_i+d_i}$+$\dfrac {1} {\sum_{i=1}^{n}(\dfrac {1}{d_i})+1}$=$S$. Let $x=\sum_{i=1}^{n}\dfrac {1}{d_i}$ We can get $(c_i+d_i)(\dfrac{1}{c_i}+\dfrac {1}{x^2d_i})\ge (1+\dfrac {1}{x})^2$ from cauchy then $\sum_{i=1}^{n}\dfrac {1}{c_i+d_i}\le$$\dfrac{\sum_{i=1}^{n}\dfrac {1}{c_i}+\dfrac {1}{x^2d_i}}{(1+\dfrac {1}{x})^2}$. Then $S-\dfrac {1}{x+1}\le$$\dfrac{\sum_{i=1}^{n}\dfrac {1}{c_i}+\dfrac {1}{x^2} (x)}{(1+\dfrac {1}{x})^2}\le\dfrac {1}{1+\dfrac {1}{x}}$. Then $S\le 1$

Also P1 in https://artofproblemsolving.com/community/c6h2600772p22443122 (with induction) proves $$\frac{1}{1+a_1 + a_2 + \cdots + a_t}+ \sum_{i=1}^{t} \frac{1}{x_i + \frac{1}{a_i}} \le 1.$$

An incredible solution created by me last year. Now I completely don't know what was the motivation. $$\sum_{i=1}^{n}\frac {1}{c_i}=\sum_{i=1}^{n}\sum_{k=1}^n\frac {a_{k,i}}{c_i^2}=\sum_{k=1}^{n}\sum_{i=1}^n\frac {1}{a_{i,k}\cdot c_i^2}\ge\sum_{k=1}^{n}\dfrac{\left(\sum_{i=1}^n\frac 1{c_i}\right) ^2}{\sum_{i=1}^na_{i,k}}=\left(\sum_{i=1}^n\frac 1{c_i}\right) ^2\sum_{k=1}^{n}\frac 1{c_k}=\left(\sum_{i=1}^n\frac 1{c_i}\right) ^3.$$Therefore $\sum_{i=1}^{n}\frac {1}{c_i}\le 1.\Box$