What can we say about $S_k=\sum_{i=1}^k r_i^s$, $0Consider $0<s<1$ fixed. What can we say about
$$S_k=\inf\left\{\sum_{i=1}^k r_i^s: 0<r_i<1 \text{ and }\sum_{i=1}^k r_i=1\right\}?$$ Does it go to $\infty$ if $k \to \infty$ or can we say that it increases if $k$ increases? I am not getting any clue on what I should try.
 A: Let $\mathbf{r} = (r_1, \dots, r_k)$ denote a $k$-tuple.
Let $S(\mathbf{r}) = \sum_{i=1}^k r_i^s$.
I think that for any $k \ge 2$, the infimum of $S(\mathbf{r})$ over all possible $\mathbf{r}$ is $1$.
So let $k$ be given, and let $\delta>0$ be some small number. We want to show that we can find a $k$-tuple $(r_1, \dots, r_k)$ such that $S(\mathbf{r}) \le 1 + \delta$.
Let $r_1 = (1 - \epsilon)$, and let $r_2 = r_3 = \dotsm = r_k = \epsilon / (k-1)$. Here, $\epsilon$ is a parameter that we will determine later.
Then the $r_i$'s sum to $1$, and
$$ S(\mathbf{r})
= (1-\epsilon)^s + k\left(\frac{\epsilon}{k-1}\right)^s
\le 1 + k\left(\frac{\epsilon}{k-1}\right)^s
\le 1 + \delta
$$
In order to make the last inequality valid, we just choose $\epsilon$ super small so that $k\left(\frac{\epsilon}{k-1}\right)^s \le \delta$, which is clearly possible.
A: Hint :
We can use Jensen's inequality and get :
$$\sum_{i=1}^{k}r_i^s\leq k\Big(\frac{\sum_{i=1}^{k}r_i}{k}\Big)^s=k^{1-s}$$
On the other hand putting $r_i=\frac{1}{y_i}$ with Jensen's inequality again :
$$k\Big(\frac{\sum_{i=1}^{k}\frac{1}{r_i}}{k}\Big)^{-s}=k\Big(\frac{\sum_{i=1}^{k}y_i}{k}\Big)^{-s}\leq \sum_{i=1}^{k}y_i^{-s}=\sum_{i=1}^{k}r_i^s$$
