Restricted character $\chi_H \in \operatorname{Irr}(H)$ implies $\chi \in \operatorname{Irr}(G)$ I'm reading Isaacs' Character Theory of Finite Groups, on page 26 it states the following fact:

Let $H \subseteq G$ be groups and let $\chi_H$ be the restriction of a
character $\chi$ of $G$. If $\chi_H$ is irreducible then so is $\chi$.

This is just stated without proof so I think it's probably something really obvious, however I'm having trouble proving it.
Using the fact that $(\chi_H, \chi_H)_H = 1$ I found $(\chi, \chi)_G = \frac{|H|}{|G|} + \frac{1}{|G|}\sum_{g \notin H} \chi(g)\overline{\chi(g)}$ but I seem to be stuck here, I would appreciate any hint on how to proceed!
 A: Hint: Use a proof by contradiction. Assume $\chi$ is reducible.
Here is the character theory proof:

 If $\chi = \theta+\phi$, then $\chi_H = \theta_H + \phi_H$ and $[\chi_H,\chi_H] = [\theta_H,\theta_H] + 2 Re([\theta_H,\phi_H]) + [\phi_H,\phi_H]$. If $\theta$ and $\phi$ are (nonzero) characters, then so are $\theta_H$ and $\phi_H$, so $[\theta_H,\theta_H] \geq 1$ and $[\phi_H,\phi_H] \geq 1$ and $[\theta_H,\phi_H] \geq 0$. This gives $[\chi_H,\chi_H]\geq 2$, a contradiction.

Here is a representation proof (you don't need to know it now, but you will want to know. This is from chapter 1, page 10. One of the coolest parts of ordinary character theory is 99% of it doesn't need chapter 1.).

 If $\chi$ is not irreducible, then the associated representation has a basis so that every matrix $X(g)$ is block upper triangular with zeros in the bottom $k \times (n-k)$ corner, where $n=\chi(1)$ and $k=\theta(1)$ and $n-k = \phi(1)$. This is the definition of reducible. However, the definition of $H$-reducible is the same, except we only need the matrices $X(h)$ to have that bottom corner be zero. Since $H \subseteq G$, it is clear that being $G$-reducible is stronger then (so implies) $H$-reducible.

Intuition
To me, part of the magic of character theory is how the same number counts completely different things. In your attempt, you have used the "sum of products of traces, divided by the group order" (definition) of the inner product. It is amazing that even comes out to be a rational number, much less an integer (Cor 2.17).
For example, the simple group of order 60 has a character $\chi$ and element $g$ with $\chi(g) = \dfrac{1+\sqrt{5}}{2}$, so $\chi(g) \overline{\chi(g)} = |\chi(g)|^2 = \dfrac{3+\sqrt{5}}{2}$. Divide that by 60, and add up some similar things to get... 1. Chapter 3 focuses on some ways to use this pretty amazing coincidence. Burnside's $p^a q^b$ theorem is little more than this (Theorem 3.9 and induction).
Inner product $[-,\chi_i]$ as counter
However, the inner product counts something else as well. By Theorem 2.8, every class function $\phi$ is of the form $\sum a_i \chi_i$ where the sum is over the irreducible characters $\chi_i$, and $a_i \in \mathbb{C}$. From the first orthogonality relation (Theorem 2.14) especially expressed in the properties on the top of page 21, we can compute $a_j$ as:
$$[\phi,\chi_j] = \left[\sum a_i \chi_i, \chi_j \right] = \sum a_i [ \chi_i, \chi_j ] = \sum a_i \delta_{ij} = a_j$$
I recommend thinking of the inner product in this second way: $[-,\chi_j]$ counts how many times $\chi_j$ appears when expression $\phi$ as a sum of irreducibles. The inner product is how many times the irreducible is in the reducible.
Inner product $[\chi,\chi]$ as Euclidean length squared
Now $[\chi,\chi]$ is a little more complicated to reason about in general, so I don't recommend it as your primary concept. It is a good thing, but not the primary thing. (If you know numerical linear, then this is a 2-norm, but counting is a 1-norm, and it can be confusing to mix them.)
If $\chi = \sum a_i \chi_i$, then $[\chi,\chi] = \sum |a_i|^2$. It is the square of the "length" of the vector. That is similar to counting up the pieces, but not exactly the same.
If $(\chi_i)_H = \sum b_{ij} \phi_j$ for $\phi_j$ the irreducibles of $H$, then $$[\chi_H,\chi_H] = \left[ \sum a_i (\chi_i)_H, \sum a_j (\chi_j)_H \right] = \sum_{i,j} a_i \bar{a_j} \left[ \sum_k b_{ik} \phi_k, \sum_l b_{jl} \phi_l \right] = \sum_{i,j,k,l} a_i \overline{a_j} b_{ik} \bar{b_{jk}} \delta_{lk}$$
$$= \sum_{i,j,k} a_i \overline{a_j} b_{ik} \overline{b_{jk}} = \|B\vec{a}\|_2^2$$
Comparing $[\chi_H,\chi_H]$ to $[\chi,\chi]$ can be a little complicated in general, though I suspect $[\chi_H,\chi_H] \geq [\chi,\chi]$ is quite common.
Counterexample for the inequality $[\chi_H,\chi_H] \geq [\chi,\chi]$ for class functions
However, it need not hold for all class functions. For example, take $G$ to be non-abelian of order 6, and $H$ to be its subgroup of size 3. $G$ has two distinct linear characters, $\chi_1$ and $\chi_2$, but on restriction to $H$ they are equal (to the trivial character, $\chi(h)=1$). Define the class function $\chi= \chi_1 - \chi_2$. Then $[\chi,\chi]=1^2+(-1)^2=2$, but $[\chi_H,\chi_H] =[0,0]=0$. (This is a very silly way to use chapter 7.)
TODO: Settle the case for characters
I got distracted by the class function case and haven't had time yet to handle the character case. I think it likely holds there, but I've sufficiently worried myself with complex numbers that I've forgotten how to count with non-negative integers :-)
