Verifying the set proof

Question

$$ x_1 + x_2 + \dots + x_{|T|} > 2|T|\overline x $$ Rearranging and substituting for $|S|$ \begin{align} & x_1 + x_2 + \dots + x_{|T|} > \frac{2|T|}{n} (x_1 + \dots + x_n) \\ & x_1 + x_2 + \cdots + x_{|T|} > \frac{2|T|}{|S|} (x_1 + \cdots + x_n) \\ & \frac{|S|}{2} \left( \frac{x_1 + \dots x_{|T|}}{x_1 + \cdots + x_n} \right) > |T| \end{align} If we look at $\frac{x_1 + \cdots +x_{|T|}}{x_1 + \cdots + x_n}$, we can see that if $|T| < n$, then this number lets call it $k$ is less than one, if $|T| = n$, then it is one. Since we assumed that $n \geq |T| \geq \frac{|S|}{2}$ we can see that: \begin{align} & \frac{|S|}{2} \left(\frac{x_1 + \dots x_{|T|}}{x_1 + \dots + x_n}\right) > |T| \\ & \frac{|S|}{2} \times k > |T| , k \leq 1 \end{align} If $k = 1 (|T| = n)$, we can see that $\frac{|S|}{2} > |T|$, which is a contradiction since we assumed $\frac{|S|}{2} \leq |T|$. If $k < 1, (\frac{n}{2} \leq |T| < n)$, then we can see that $k' > |T|$ where $k' < \frac{|S|}{2}$, since $k' = \frac{|S|}{2} \times k, k < 1$. So, $\frac{|S|}{2} > \frac{|S|}{2} \times k > |T|$. This is also a contradiction, so we can see that when when $|T| \geq \frac{n}{2}$ we arrive at a contradiction.

Answer 1

Your argument is unnecessarily complex.

Suppose that $\#T≥\frac {\#S}2$. Then consider $$\sum_{x\in T}x> 2\overline x\frac {\#S}2=\overline x\,n$$

But $\overline x\,n=x_1+\cdots+x_n$ hence some element in $S-T$ must be $<0$, contradicting the assumption that all the $x_i$ were non-negative.