KöMaL 2018 A721

Let $n\ge 2$ be a positive integer, and suppose $a_1,a_2,\dots,a_n$ are positive real numbers whose sum is $1$ and whose squares add up to $S$. Prove that if $b_i=\frac{a_i^2}{S}$ ($i=1,\dots,n$), then for every $r>0$, we have $\sum_{i=1}^n \frac{a_i}{{(1-a_i)}^r}\le \sum_{i=1}^n \frac{b_i}{{(1-b_i)}^r}.$