show that there exists a set $S$ satisfying an inequality 
In the solution to the problem shown below, why can they assume WLOG that $s_{i_1} + s_{i_2} \ge -(t_{i_1} + t_{i_2})$? I tried proving the result in the case where this doesn't hold by showing that $-(s_{i_1} + s_{i_2}) + t_{i_1} \leq -s/6$ or $-(s_{i_1} + s_{i_2}) + t_{i_2})\leq -s/6$ (which would hold if their sum was at most $-s/3$) but I wasn't able to make much progress.


Also when they set $r_1 = r_2 = r_3 = 1, r_4 = r_5 = r_6 = -1,$ why is $\frac{1}6$ the best possible value for the bound?



 A: 
In the solution to the problem shown below, why can they assume WLOG that $s_{i_1} + s_{i_2} \ge -(t_{i_1} + t_{i_2})$?

Because if the theorem to be proved works for $\{r_k\}_{k=1}^n$, it will also work for $\{-r_k\}_{k=1}^n$. Negating the $r_k$ swaps the values of $s_i$ and $t_i$ for each $i$.
So if it turns out that you cannot pick $i_1, i_2$ so that $s_{i_1} + s_{i_2} \ge -(t_{i_1} + t_{i_2})$, then you can prove the result for $\{-r_k\}_{k=1}^n$ instead. And for that sequence $s_{i_1} + s_{i_2} \ge -(t_{i_1} + t_{i_2}), s_{i_1} + s_{i_2} \ge \frac s3$ will be true for some $i_1, i_2$, exactly because it wasn't ever true for the original sequence. And the truth of the theorem for this sequence immediately implies it was true for the original.

Also when they set $r_1 = r_2 = r_3 = 1, r_4 = r_5 = r_6 = -1,$ why is $\frac{1}6$ the best possible value for the bound?

The condition requires you to pick at most two elements from $r_1, r_2, r_3$, and at least one element from $r_4, r_5, r_6$. That means the highest $\sum_{i\in S} r_i$ can be is $1 + 1 - 1 = 1$. And similarly the condition requires you to pick at least one element from $r_1, r_2, r_3$ and at most two elements from $r_4, r_5, r_6$, so the lowest $\sum_{i\in S} r_i$ can be is $1 - 1 - 1 = -1$.
So for any pick of index set $S$ meeting the condition, $\left|\sum_{i\in S} r_i\right| \le 1 = \frac 16\sum_{i=1}^n |r_i|$.
