lemma 1 $max(xk,yz,yk) = l$ then $$xk+yz \leq xz+l$$ proof if $xk =l$ then $y\leq x$ so $xk+yz- xz-l=-(x-y)z\leq 0$ if $yk =l$ then $x\leq y$ and $z\leq k$ so $xk+yz- xz-l=-(y-x) (k-z)\leq 0$ if $yz =l$ then $k\leq z$ so $xk+yz- xz-l=-x (k-z)\leq 0$ let prove that if $a_1, \ldots , a_n$,$b_1, \ldots , b_n$ positive numbers $r_k$ maximum of the products $a_ib_j$ over the sets $max(i, j) = k$ then $(a_1 + \ldots + a_n) (b_1 +\ldots + b_n) \leq n (r_1 + \ldots + r_n)$ with induction n=2 from lemma 1 true true for $1 , \ldots ,n-1$ for n=n from lemma 1 we get for $i=2 , \ldots ,n$ $a_1b_i+b_1a_i\leq a_1b_1+r'_i$ and $r'_i$ is maximum of the products $a_ib_1,a_1b_j,a_ib_i$. from induction $(a_2 + \ldots + a_n) (b_2 +\ldots + b_n) \leq (n-1) (m'_2 + \ldots + m'_n).$. and $m'_k$ be the maximum of the products $a_ib_j$ over the sets $(i, j)$ for which $max(i, j) = k$ and $2\leq i ,j$ .and $a_1b_1=m_1$. let add this inequalities $(a_1 + \ldots + a_n) (b_1 +\ldots + b_n) \leq nm_1+ (r'_2 + \ldots + r'_n)+(n-1) (m'_2 + \ldots + m'_n)\leq n (r_1 + \ldots + r_n).$ we get. $(a_1 + \ldots + a_n) (b_1 +\ldots + b_n) (c_1 +\ldots + c_n) \leq n (r_1 + \ldots + r_n)(c_1 +\ldots + c_n) \leq n^2 (m_1 + \ldots + m_n)$ we get

I wonder if all this kind of problems can be done by induction (2024-TST-18???) We first prove the (2D) version with only $a_i,b_j.$ Define $l_k:=\max_{\max{i,j}=k}a_ib_j,$ then $$(a_1+a_2+\cdots +a_n)(b_1+b_2+\cdots +b_n)\le n(l_1+l_2+\cdots +l_n).$$Proof. induct on $n.$ Base case is trivial. Now inequality holds when $n-1,$ we only need $$b_n(a_1+a_2+\cdots +a_{n-1})+a_n(b_1+b_2+\cdots +b_{n-1})+a_nb_n\le l_1+l_2+\cdots +l_{n-1}+nl_n.$$$$\iff\sum_{k=1}^{n-1}(a_nb_k+a_kb_n-l_k-l_n)+(a_nb_n-l_n)\le 0.\Box$$Now $\Sigma a_k\Sigma b_k\Sigma c_k\le n\Sigma l_k\Sigma c_k\le n^2\Sigma m_k,$ done$.\Box$