I'm taking a course in quadratic forms at the moment. There is no text book or notes so all I have to go from is what the teacher writes on the blackboard. We had the following lemma some weeks ago: $ \newcommand{\F}{F}\newcommand{\K}{K} $
Lemma: Let $ \F $ be a formally real field and $ \K = \F(\sqrt{a}) $ a quadratic extension. Then $ \K $ is not formally real if and only if $ -a \in \Sigma\F\backslash\{0\} $ where $ \Sigma\F = \{a\in \F\mid a\ \text{is a sum of squares}\} $.
I'm pretty sure the proof went something like this:
($ \Leftarrow $) If $ -a \in \Sigma\F\backslash\{0\} $ then $ (\sqrt{a})^2 - a = 0 $ implies $ (\sqrt{a})^2 +\sum_i b_i^2 = 0 $. Thus $$ 1 + \sum_i \left(b_i/\sqrt{a}\right)^2 = 0 \Leftrightarrow -1 = \sum_i\left(b_i/\sqrt{a}\right)^2, $$ which means $ -1 $ is a sum of squares in $ \K $ and thus $ \K $ is not formally real.
($ \Rightarrow $) If $ -1 = \sum_i (b_i+c_i\sqrt{a})^2 $, $ b_i,c_i\in\F $ then $ \sum_i b_i^2 + a\sum_ic_i^2 = -1 $ and thus $ (1+\sum_ib_i^2)(\sum_ic_i^2)^{-1} = -a $. As $ \Sigma\F\backslash\{0\} $ is a group, this implies $ -a \in \Sigma\F\backslash\{0\} $ since both $ 1+\sum_ib_i^2, \sum_i c_i^2 \in \Sigma\F\backslash\{0\} $. Note that $ \sum_ic_i \neq 0 $ as $ \F $ is formally real.
Note the part in boldin the second half of the proof: How is $ \Sigma\F\backslash\{0\} $ a group?
I can see that it must be a multiplicative group and I know that the product of two sums of squares is itself a sum of squares, but how do I have inverse elements?
You have inverses, because
$$\frac{1}{\sum_{i=1}^n c_i^2} = \frac{\sum_{k=1}^n c_k^2}{\left(\sum_{i=1}^n c_i^2\right)^2} = \sum_{k=1}^n \left(\frac{c_k}{\sum_{i=1}^n c_i^2}\right)^2$$
if $\sum_{i=1}^n c_i^2 \neq 0$.