Let $H$ be a subgroup of $G$ and $\chi_1, ... \chi_r$ irreducible characters of G. Let $\psi$ be an irreducible character of $H$ and $d_1, ... , d_s$ integers such that $\psi \uparrow G = d_1\chi_1 + ... + d_k\chi_k$. Show that $$\sum_i d_i^2 \leq |G:H|$$
We have $|G:H|\psi(1) = d_1\chi_1(1) + ... + d_k\chi_k(1)$, where $d_i = \langle \psi \uparrow G, \chi_i \rangle_G = \langle \psi, \chi_i \downarrow H \rangle_H$ by Frobenius Reciprocity Theorem.
Hence, since $\psi$ is irreducible, $\chi_i \downarrow H = d_i \psi + \beta$ where either $\beta$ is a character of $H$ or $\beta=0$.
Thus $\chi_i(1) \geq d_i \psi(1)$ and therefore $$|G:H|\psi(1) \geq (d_1^2 + ... + d_k^2)\psi(1)$$
What I'm stuck on
I don't understand the bold italic line in the proof, I get that since in H $\psi$ is irreducible, the $\chi_i \downarrow H$ must include it some number of times each, adding up to 1 (I think), but how do you know it is $d_i$ each time? How do you know $\beta$ is non-negative?