$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$ for natural numbers $n_1 < n_2 < ... < n_k$ Prove that for natural numbers $n_1 < n_2 < ... < n_k$:
$$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$$
I have no idea how to do that. I would go for induction, but there is an infromation that $n_1 < n_2 < ... < n_k$ and induction would not recognise that fact.
What's funny is that I think I saw once that exercise on stack exchange but wasn't able to find it this time.

We can go further to even state the biconditional equality condition. The following is a quote of Proposition 3 from Sum of Cubes is Square of Sum by Barbeau and Seraj, with permission from me, one of the authors:

We prove by induction that:
For integers $a_k$, if $1 \leq a_1 < a_2 < \cdots < a_n$ then
$$a_1^3 + a_2^3 + \cdots + a_n^3 \ge (a_1 + a_2 + \cdots + a_n)^2,$$
with equality if and only if $a_k = k$ for $1 \le k \leq n$.
This is clear for $n = 1$. Assume that the proposition holds for some natural $n$; we will prove it for $n+1$. Note that $a_k \ge k$ and that $a_{n+1} - k \ge a_{n+1-k}$ for all values of $k$. We begin with the fact that
$$(a_{n+1} - n)(a_{n+1} - n - 1) \ge 0$$
with equality if and only if $a_{n+1} = n+1$. Then expanding gives
\begin{align*}
 a_{n+1}^2 - a_{n+1} & \ge 2na_{n+1} - n(n+1) = 2 \left[ na_{n+1} - {{n(n+1)}\over 2} \right]\\
 & = 2 \left( \sum_{k=1}^n (a_{n+1} - k) \right) \ge 2 \sum_{k=1}^n a_{n+1-k} = 2 \sum_{k=1}^n a_k,
\end{align*}
with equality if and only if $a_k = k$ for all $k$. By the induction hypothesis, we have that
\begin{align*}
 \sum_{k=1}^{n+1} a_k^3 &=  a_{n+1}^3 + \sum_{k=1}^n a_k^3 \ge a_{n+1}^2 + 2a_{n+1}\left(\sum_{k=1}^n a_k \right) + \left(\sum_{k=1}^n a_k \right)^2\\
 & = \left(a_{n+1} + \sum_{k=1}^n a_k \right)^2 = \left( \sum_{k=1}^{n+1} a_k \right)^2,
\end{align*}
as desired.

