Inequality for Character Degrees of finite groups

Question

Let $G$ be a finite group, and let $d_1,d_2,\dots,d_n$ be the dimensions of the irreducible representations. It is well-known that $\sum_{i=1}^n d_i^2=|G|$. If I am not mistaken, one has the following inequality $$ \sum_{i=1}^n d_i^{3/2}\geq \sqrt{|G|}. $$ Is this known or obvious (or false)? For abelian groups the inequality is far from sharp, because we can replace the right hand side with $|G|$. But for the symmetric group it seems pretty sharp. For example, for $S_6$, the left-hand side is 27.2688 and the right-hand side is $26.8328$.

Answer 1

Never mind. The inequality that I wrote down is obvious, because $$ (\sum_{i=1}^n d_i^{3/2})^2\geq \sum_{i=1}^n d_i^3\geq \sum_{i=1}^n d_i^2=|G|. $$ I accidentally calculated $\sum_{i=1}^n \sqrt{d_i}$ on the left-hand side. So it might be an interesting question whether $$ \sum_{i=1}^n \sqrt{d}_i\geq \sqrt{|G|}. $$ I'm pretty sure this fails for large symmetric groups, because the left-hand side is less or equal than $p(n)|S_n|^{1/4}$ where $p(n)$ is the number of partitions. $p(n)$ grows subexponentially, whereas $|S_n|$ grows superexponentially. So we have $p(n)<|S_n|^{1/4}$ for large $n$. So the left-hand side will eventually be smaller than $\sqrt{|S_n|}$.

Answer 2

Let $G$ have $k$ conjugacy classes. Then Cauchy Schwarz seems to give $\sum_{i=1}^{k} \sqrt{d_i} \leq \sqrt{k}\sqrt{\sum_{i=1}^{k} d_i }$ and this (using C-S) again is at most $k^{\frac{3}{4}} |G|^{\frac{1}{4}}$. So as long as $k^{3} < |G|,$ the inequality you state is violated. There are many groups $G$ for which $k^{3} < |G|.$ One example is the alternating group $A_{6}$ which has $7$ conjugacy classes and order $360 >7^{3}.$