Here's a proof of 11.3. We induct on $n$. Note that for $k=1$, $a_1\ge 1$ so $a_1^3\ge a_1^2$ trivially. Suppose that for any $a_0,a_1,\dots,a_{n-1}$ with $a_k\ge a_{k-1}+1$ the conclusion holds. We take $a_0,\dots,a_n$ and observe \[ \left(\textstyle\sum_{1\le k\le n}a_k\right)^2 = a_n^2 + 2a_n \textstyle\sum_{1\le k\le n-1}a_k + \left(\textstyle\sum_{1\le k\le n-1}a_k\right)^2. \]As \[ \textstyle\sum_{1\le k\le n-1}a_k^3 \ge \left(\textstyle\sum_{1\le k\le n-1}a_k\right)^2, \]it suffices to prove \[ a_n^2 \ge a_n + 2\textstyle\sum_{k\le n-1}a_k. \]Now, we have $a_{n-1}\le a_n-1$, $a_{n-2}\le a_{n-1}-1\le a_n-2$, and so on. Thus, $\textstyle 2\sum_{k\le n-1}a_k \le 2(n-1)a_n -n(n-1)$. With this, it suffices to prove $a_n^2 -(2n-1)a_n + n(n-1)\ge 0$, i.e., $(a_n-n)(a_n-(n-1))\ge 0$. As $a_1\ge a_0+1=1$, $a_2\ge a_1+1\ge 2$, we have $a_n\ge n$, so this inequality indeed holds.