The sum of the largest $n$ integers is greater than the sum of the remaining $n+1$ Let $a_1\geq a_2\geq\ldots\geq a_{100n}$ be positive integers such that if we choose any $2n+1$ numbers from them, the sum of the largest $n$ is strictly greater than the sum of the remaining $n+1$. Prove that

$$(n+1)(a_1+\ldots+a_n)&gt;a_{n+1}+\ldots+a_{100n}.$$

For the case $n=1$, this becomes $a_1\geq\ldots\geq a_{100}$, and we need to prove $2a_1&gt;a_2+\ldots+a_{100}$. We have $a_1&gt;a_2+a_3, a_2&gt;a_3+a_4, \ldots$, and it's not clear how the desired inequality follows.
$$
2a_1 &gt; 2a_2+2a_3&gt;a_2+2a_3+a_3+a_4 = a_2+a_3+a_4+2a_3
$$
Similar for $a_3$ and you will gain desired inequality. Then just use induction.
$$
2a_1 &gt; 2a_2+2a_3&gt;a_2+2a_3+a_3+a_4 = a_2+a_3+a_4+2a_3
$$
Similar for $a_3$ and you will gain desired inequality. Then just use induction.