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which satisfy 
![$ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$](http://latex.artofproblemsolving.com/c/d/a/cdae05ad34239366b0dd3921243522c711b6ae58.png)
method, I can assum 
![$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $](http://latex.artofproblemsolving.com/c/d/4/cd481672bee1e87d8c97ac623436e0ec75cfa73a.png)
are given such that 
. Prove the inequality 
, let 
denote the smallest possible value of 
where 
are sets such that 
and 
whenever 
. Determine for each positive integer .
and 
are such that 
as sum of the 
-th row's entries and 
as sum of the 
-th column's entries, find the maximum sum of the entries on the main diagonal.
and are such that as sum of the -th row's entries and as sum of the -th column's entries, find the maximum sum of the entries on the main diagonal.
is an upper bound; the question is whether we can get there. First, we should note that the maximum exists: the set of such matrices is closed, and the trace function is continuous so it achieves its maximum. Now suppose we have one of these matrices of trace strictly less than 
; we'll show by an old trick that it's not of maximal trace, and this finishes the result. Let 
be our matrix. Since 
there must be some such that 
. By the defining conditions of our set of matrices, it follows that there are and 
, neither equal to , such that 
and 
. But then we can replace by a new matrix in which we increase 
and 
and decrease 
and 
all by the same (positive) amount while keeping all entries nonnegative. This resulting matrix has trace strictly larger than , as desired.
in natural numbers m and n.
