Sum and minimum in $3\times 3$ table A $3\times 3$ table is filled with non-negative real numbers so that each row sums to $1$. A subset of cells is called admissible if it has size $3$ and the three cells all lie in different rows and columns. For each admissible set $A$, let $s(A)$ be the sum of the numbers in the three cells and $m(A)$ be the minimum of those numbers.
There must be an admissible set $A$ with $s(A)\geq 1$. (Just take three disjoint admissible sets; since the total sum is $3$, one set must have sum at least $1$.) Is it true that there is always an admissible set $A$ such that $s(A)\geq 1$, and $m(A)$ is the highest among all admissible sets?
 A: Yes, this is true. Take $A$ to be an admissible set achieving the highest possible $m(A),$ but subject to that condition, with the minimum achievable number of cells equal to $m(A).$ (So we prefer one entry to equal $m(A)$ rather than two.)
Without loss of generality, $A$ is the set of diagonal cells, and the $(1,1)$ cell is $m(A).$ Consider which cells are not in $A$ but are strictly greater than $m(A).$ Call these "augmenting".
If there is a row with no augmenting cells, then the element of $A$ in that row is at least $1-2m(A),$ which gives $s(A)\geq 1.$ So we are done unless each row has an augmenting cell.
If the $(2,1)$ and $(3,1)$ cells are not augmenting, then the total of the cells in the $(2,2),(2,3),(3,2),(3,3)$ positions is at least $2-2m(A).$ Therefore either of $\{(1,1),(2,2),(3,3)\}$ or $\{(1,1),(2,3),(3,2)\}$ has sum at least $1,$ and by the previous paragraph these both have minimum at least $m(A),$ so we're done.
The remaining case is that one of the $(2,1)$ or $(3,1)$ cells is augmenting. Some cell in the first row is also augmenting. If $(1,2)$ and $(2,1)$ are augmenting then the set $(1,2),(2,1),(3,3)$ is an admissible set contradicting the choice of $A$ - it either has a higher minimum or the same minimum but fewer elements achieving the minimum. Similarly we are done if $(1,3)$ and $(3,1)$ are augmenting. Otherwise wlog $(2,1),(1,3)$ are augmenting and $(3,1),(1,2)$ are not (the other case just swaps $2$ and $3$). Then $(3,2)$ must be augmenting because it is the only remaining cell in the third row, so $(2,1),(1,3),(3,2)$ forms an admissible set with higher minimum.
