![n \geq 4](/media/m/4/f/2/4f2c159d62a3247dd33ee8d022ae9f37.png)
players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players
![bad](/media/m/0/a/a/0aa472cd535db3abac88a27108fcd7e0.png)
if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that there is no bad company in this tournament. Let
![w_i](/media/m/f/6/7/f678b0a6ed723dbe29fe5d06f491f65e.png)
and
![l_i](/media/m/b/2/6/b2647bded77d01395618ce39635fe411.png)
be respectively the number of wins and losses of the
![i](/media/m/3/2/d/32d270270062c6863fe475c6a99da9fc.png)
-th player. Prove that
![\sum^n_{i=1} \left(w_i - l_i\right)^3 \geq 0.](/media/m/4/a/9/4a9f1333529a5f11bf615870e4bc2700.png)
Proposed by Sung Yun Kim, South Korea
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$n \geq 4$ players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players $bad$ if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that there is no bad company in this tournament. Let $w_i$ and $l_i$ be respectively the number of wins and losses of the $i$-th player. Prove that
$$\sum^n_{i=1} \left(w_i - l_i\right)^3 \geq 0.$$
Proposed by Sung Yun Kim, South Korea