TITLE: Maximizing speed and fitness with fewest players
QUESTION [22 upvotes]: There are $n$ football players, each of whom has a speed $s_i\in[0,1]$ and fitness $f_i\in[0,1]$. The sum of the speeds of all players is $1$, and the same is true for fitness. We want to choose a subset of players so that the sum of speeds and the sum of fitnesses are both at least $1/2$. Let $a$ be the size of the smallest such subset.
Suppose we perform the following "greedy" algorithm: keep picking players with the maximum sum $s_i+f_i$, until either we have satisfied $\sum s_i\geq 1/2$ or $\sum f_i\geq 1/2$, and then pick the players with the maximum possible attribute that we haven't satisfied yet. Let $b$ be the size of the set we get. (Ties are broken arbitrarily.)
Is it true that $b/a\leq 3/2$ always? It is possible that $b/a=3/2$, as shown by the example where $n=3$, $(s_i)=(0.4,0.6,0)$ and $(f_i)=(0.4,0,0.6)$. The minimum-size subset is $2$, by picking the 2nd and 3rd players, but the algorithm picks all three players. On the other hand, it is not hard to show that $b/a\leq 2$ must hold (e.g. following Alex Ravsky's argument)

REPLY [2 votes]: Here is a construction with $b/a=2-\tfrac{2}{k+2}\approx 2$ for any integer $k\geq 2.$ So the bound $b/a\leq 2$ is the optimum bound independent of $n.$ Take these categories of players:

one player with $(s_i,f_i)=(\tfrac12-\tfrac1{2k},0)$
$k$ players with $(s_i,f_i)=(0,\tfrac1{2k})$
$k+1$ players with $(s_i,f_i)=(\tfrac1{2k(k+1)},\tfrac1{2(k+1)})$ - note $s_i+f_i=\tfrac1{2k}$ here
$k(k+1)$ players with $(s_i,f_i)=(\tfrac1{2k(k+1)},0)$

The optimum is achieved by taking the players from category 1 and 3, totalling $k+2.$ The greedy algorithm will take the player from category 1 but there is then a tie for $s_i+f_i$ between 2 and 3 and we can choose to first take all $k$ players from category 2. This achieves a total fitness of $\tfrac 1 2$ so the remaining players are chosen based on speed and we can take any $k+1$ players from category 3 or 4. This gives a total of $2k+2$ for the greedy algorithm.