Has it been proven that the sum of powers is greater than the power of the sum? I apologize if the title is confusing, but I essentially just want to know if there exists an accepted, rigorous proof of the inequality discussed in this question.
It seems as though it would have to be true that given $k \geq 1$ and $a_{i} \geq 0$, we'll have that:
$\sum_{i=0}^{n}a_{i}^{k} \leq (\sum_{i=0}^{n}a_{i})^{k}$, 
but I'm having trouble coming up with a definitive proof myself that works for any $k \geq 1$, regardless of whether it's an integer or not, and I haven't been able to find any published proofs for this inequality despite spending a significant amount of time searching for one.
The accepted answer to the question I linked above seems completely reasonable, but it's certainly not rigorous. Add to this the fact that I can't find a rigorous proof of this inequality, or even any discussion of it whatsoever for that matter, anywhere else on the internet, despite spending over an hour reading through the articles for every significant and/or named inequality on wolfram mathworld and wikipedia, and my confidence that this inequality is in fact valid is not very high.
Can anyone provide any insight into this, or point me to a source that says something definitive about this inequality one way or the other? I'm especially interested in the cases for which $k$ is not an integer.
 A: Let $\lambda = \sum\limits_{i=0}^n a_i$. 
If $\lambda = 0$, all $a_i = 0$ and the inequality is trivially true. 
WOLOG, let's assume some $a_i > 0 \implies \lambda > 0$.
For any $i = 0, 1,\ldots, n$, let $\displaystyle\;b_i = \frac{a_i}{\lambda}$, we have $\;b_i \in [0,1] \implies b_i^k \le b_i$.
Together with $\sum\limits_{i=0}^n b_i = 1$, we find
$$\sum_{i=0}^n a_i^k = \lambda^k \sum_{i=0}^n b_i^k \le \lambda^k \sum_{i=0}^n b_i = \lambda^k = \left(\sum_{i=0}^n a_i\right)^k$$
A: If you write $$f(x_1,x_2,\dots,x_n)=\left(\sum{x_i}\right)^k-\sum{x_i^k},$$ then $f(0,0,...,0)=0$ and it's easy to show that all the first-order partial derivatives $\frac{\partial f}{\partial x_j}$ are strictly positive when $k>1.$  Indeed,$$
\frac{\partial f}{\partial x_j}=k\left(\left(\sum{x_i}\right)^{k-1}-x_j^{k-1}\right)>0$$ when $x_i>0 \forall i$.
Because $g(x)=x^{k-1}$ for $k>1, x>0$ is an increasing function and $\sum x_i > x_j\; \forall j$ since $x_i>0\;\forall i$
A: If you can prove $a^k+b^k\le(a+b)^k$ for $a$, $b>0$, then the general case
follows by induction. This is the same as $t^k+(1-t)^k\le1$
for $0<t<1$, taking $t=a/(a+b)$, By calculus, $t^k+(1-t)^k$
has only one extremum in $(0,1)$, at $t=1/2$, and that's a minimum when $k>1$. So on $[0,1]$, $t^k+(1-t)^k\le1$.
