Quote: Looking for the ISL 2006? Check this out! Let $A_{1}$ be the best and $A_{10}$ the worst team. and let $f(A_{1},...,A_{1}0)= x_{1}+2x_{2}+...+10x_{10}$ In a game that $A_{j}$ wins over $A_{i}$, $i < j$ , magicly changing the fate of the game, $f$ decreases.So minimum occures when for all the games $A_{i}$ wins over $A_{j}$ $i < j$. hence \[f \geq 9+2(8)+3(7)+...+8(2)+9 =165 \]

Isn't this direct consequence of rearrangement inequality?

ok here's a double counting approach it is obvious that the sum of points of every k persons is at least $\binom{k}2$ then the sum \[(x_1+x_2+x_3+\cdots+x_{10})+(x_2+x_3+\cdots+x_{10})+\cdots+(x_9+x_{10})+x_{10} \ge \binom{10}2+\binom{9}2+\cdots+\binom{2}2=165\]\[\color{magenta} \huge{\boxed{\mathbf{Q.E.D}}}\]