Let $A_n$ be the alternating group of degree $n$ and $b_n$ be the maximal character degree of $A_n$. I.E. $b_n=\mathrm{max}\{\chi(1) : \chi \in \mathrm{Irr}(A_n)\}$. I want to know if there is a way to provide an upper bound on $b_n$ inductively. In particular, I am trying to prove that $b_n<nb_{n-1}$. This is at least true for small $n$ but I can't seem to find a general proof.
I have just found the following GroupProps article which claims that in general $b(G)\leq |G:H|b(H)$ where $b(G)=\mathrm{max}\{\chi(1) : \chi \in \mathrm{Irr}(G)\}$ and $H$ is a subgroup of $G$. Thus, my desired inequality follows immediately upon proving that equality cannot be attained. However, I am still unsure of how to prove this.
On a related note, we can obtain $b_n^2<\frac{|A_n|}{2}$ but I am wondering if there is some lower bound on $b_n$. In particular, is it possible to prove $b_n^2>\frac{|A_n|}{2(n+1)}$ which would give us the strict inequality that I want above.
Let $\chi$ be an irreducible character of $A_n$, and restrict $\chi$ to $A_{n-1}$. Let $\phi$ be a constituent of the restriction. By Frobenius reciprocity, $\chi$ is a constituent of the induction of $\phi$ to $A_n$, and this induced character has degree $n\cdot \phi(1)\leq n\cdot b_{n-1}$.
(As an aside, I think it is still not known what the maximal degree $b_n$ actually is for alternating groups, although there are better bounds than this in the literature.)