Understanding Terry Tao's proof of the Brauer-Fowler theorem

Question

I'm trying to understand how Terry Tao got to the last inequality in this blog post. Here is my attempt If we exclude the trivial character, then we have $$ |G|-1 = \sum_{\chi\in\hat{G}\backslash 1}{\chi(1)^2} $$ Then using Cauchy-Schwarz and a previous result $$ \sum_{\chi\in\hat{G}\backslash 1}{\chi(1)^2} \geq \frac{1}{|\hat{G}|-1}\left(\sum_{\chi\in\hat{G}\backslash 1}{\chi(1)}\right)^2 \geq \frac{1}{|\hat{G}|-1}\left(\frac{|G|}{n}\right)^2 $$ Thus, $$ |\hat{G}|-1 \geq \frac{1}{|G|-1}\left(\frac{|G|}{n}\right)^2 $$ But the inequality I want is $$ \frac{|G|-1}{n^2}\leq |\hat{G}|-1 $$ How can I arrive at this result?

Answer 1

$$ |\hat{G}|-1 \geq \frac{1}{|G|-1}\left(\frac{|G|}{n}\right)^2\implies\frac{|G|^2}{n^2\left(|G|-1\right)}\le|\hat G|-1$$

But

$$\frac{|G|^2}{n^2\left(|G|-1\right)}\ge\frac{|G|-1}{n^2}\iff|G|^2\ge(|G|-1)^2$$