Set with sum of elements equal to sum of squares The question I'm attempting to solve is

For a set $S$ such that $\sum S_i = \sum S_i^2$, which is greater: $\sum S_i^3$ or $\sum S_i^4$?

My first thought was that $$\sum S_i - \sum S_i^2 = 0 \implies \sum S_i(1-S_i) = 0$$ hints at some kind of symmetry which may be leveraged, since the question can be rephrased as proving that $\sum S_i^3(1 -  S_i)$ is less than, or greater than, $0$ for all values. Unfortunately I couldn't find a way to proceed from here.
I found the parametric solution $$a=\frac{k(k+1)}{k^2+1}$$ and $$b=\frac{k+1}{k^2+1}$$
for $a+b = a^2 + b^2$, and I could manually try some values to see which is larger, but I am looking for a proof. I tried to show $(a^4+b^4)-(a^3+b^3)$ was always on one side of $0$, but I ended up with a polynomial of degree 7, which has odd degree so must cross the x-axis! I'm guessing my work up to this point was wrong - this post is getting a bit long, but I can post if it would be useful.
I have the feeling that I am missing a very simple and intuitive proof, but it has eluded me so far.

If the $S_i$ can have any sign nothing can be said (see e.g. the example of Paul, where $\sum S_i^3 < \sum S_i^4$, whereas for $S = \{.5, .5, (1-\sqrt{3})/2\}$ the opposite inequality holds).
Hence I am assuming that $S_i \geq 0$ for every $i$.
We have that
$$
S_i^2 = (S_i)^{1/2} (S_i)^{3/2} \leq \frac{1}{2} S_i + \frac{1}{2}S_i^3
$$
hence
$$
\sum S_i^2 \leq \frac{1}{2} \sum S_i + \frac{1}{2} \sum S_i^3.
$$
Since $\sum S_i = \sum S_i^2$, it follows that
$\sum S_i^3 \geq \sum S_i = \sum S_i^2$.
On the other hand
$$
S_i^3 = S_i \cdot S_i^2 \leq \frac{1}{2} S_i^2 + \frac{1}{2}S_i^4,
$$
so that, using $\sum S_i^3 \geq \sum S_i^2$,
$$
\sum S_i^4 \geq 2 \sum S_i^3 - \sum S_i^2 \geq \sum S_i^3.
$$
