TITLE: Maximum of sum is at most the sum of maxima
QUESTION [9 upvotes]: I'd like to show the following:
$$\max_x \sum_y f(x,y) \le \sum_y \max_x f(x,y)$$
I've also been asked to show the conditions under which the two are equal.
I believe that I'm getting confused by the multiple dimensions.  Intuitively, if you take the max over $x$ of $f(x,y)$ and sum over $y$, I know that this will more than likely be greater than the prior, but I'm really stumped on this.

REPLY [10 votes]: Here's a methodical way to solve the problem.
First, you're trying to show
$$ \max_x \text{something} \le \text{something else} $$
This is equivalent to
$$ \text{for all $x$,}\quad \text{something}\le\text{something else} $$
So let $x$ be arbitrary and let's show
$$ \sum_y f(x,y) \le \sum_y \max_x f(x,y) $$
To avoid confusion, let me change the $x$ on the right, because it's not the same as the $x$ we just chose.
$$ \sum_y f(x,y) \le \sum_y \max_z f(z,y) $$
Anyway, to show $\sum_y \text{something} \le \sum_y \text{something else}$, the simplest thing that could possibly work is to show $\text{something}\le\text{something else}$ and then sum over $y$.  So let's try that: we want to show
$$ f(x,y) \le \max_z f(z,y) $$
And now... well, yes, this is true: the LHS is one of the values considered in the maximum on the RHS, so of course the RHS is larger.