Let $G$ be a finite group, and let $d_1,d_2,\dots,d_n$ be the dimensions of the irreducible representations. It is wellknown 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 lefthand side is 27.2688 and the righthand side is $26.8328$.

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 lefthand 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 lefthand 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 lefthand side will eventually be smaller than $\sqrt{S_n}$. 


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 CS) 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 $360 >7^{3}.$ 

