TITLE: Ordered field proof that squared terms are zero or positive
QUESTION [5 upvotes]: I have decided to set up a bounty for one outstanding solution to this problem. The winner with the most clear, confident, solution manual-esque response will be awarded 50 of my participation points. The answer must be clear and unambiguous. You should attempt to solve the problem as if this were a midterm/final exam environment. It must show me clearly that either my attempt was completely off or on the right track. Good luck!
I am completely lost on how to answer this question. I have a feeling this is somewhat of an awkward proof. If I am missing something all hints will help!
Prove that for all $n \in \mathbb{N}$, if $a_1 \in F,a_2 \in F,...,a_n \in F$,where $F$ is an ordered field
then $$\sum_{k=1}^{n} a_k^2 \geq 0$$
I know I have to attempt this to receive a competent answer, so here it goes:
$$\sum_{k=1}^{1} a_1^2 \geq0$$
since $|a_1|^2 \geq 0$
This looks like an induction problem but I want to argue that anything in an ordered field squared is positive or $0$.
Assume $$\sum_{k=1}^{n_0} a_k^2 \geq 0$$
Therefore 
$$\sum_{k=1}^{n_0} a_{k}^2 +a_{n_0+1}^2 \geq 0$$
Since $|a_1|^2+\ldots+|a_{n_0}|^2+|a_{n_0+1}|^2 \geq 0$
So by induction 
$$\sum_{k=1}^{n} a_k^2 \geq 0$$ for all $n \in \mathbb{N}$

REPLY [3 votes]: Definition of ordered filed
A field $(F, +, \cdot)$ together with a (strict) total order $<$ on $F$ is an ordered field if the order satisfies the following properties for all $a, b$ and $c$ in $F$:

if $a < b$ then $a + c < b + c$, and
if $0 < a$ and $0 < b$ then $0 < a\cdot b$.


Lemma:  $\forall a \in F$, $a^2 \ge 0$.
Proof: 

If $a=0$, then $a^2 = 0 \cdot (1-1)=0\cdot 1 - 0 \cdot 1 = 0-0=0$ where I have used definition of additive inverse and distributive law.
If $a>0$, then $a^2=a\cdot a>0$ by definition of ordered field.
If $a<0$, adding $-a$ to both sides, we have $0<-a$. Also $1+(-1)=0$ by squaring, $$(1+(-1))\cdot (1+(-1)) = 0$$ which can be expanded to $$1+(-1)+(-1)+(-1)^2=0$$
$$-1+(-1)^2=0$$

That is $$(-1)^2=1.$$
Hence $$a^2=1\cdot a^2 = (-1)^2 \cdot a^2 = (-a)^2>0$$
That is $$\forall a \in F, a^2 \ge 0. \square$$

Claim: $\forall n \in \mathbb{N}$, $\sum_{i=1}^n a_i^2 \ge 0.$
Base case: For $n=1$, this has been shown by the lemma.
Suppose we have $k \in \mathbb{N}$, $\sum_{i=1}^k a_i^2 \ge 0$
we want to show that $\sum_{i=1}^{k+1} a_i^2 \ge 0$.
This is true because
\begin{align}\sum_{i=1}^{k+1} a_i^2  &= \left(\sum_{i=1}^{k} a_i^2\right) + a_{k+1}^2
\ge 0\end{align}
because $\left(\sum_{i=1}^{k} a_i^2\right) \ge 0$ and $a_{k+1}^2 \ge 0$.
Hence, by mathematical induction, $$\forall n \in \mathbb{N}, \sum_{i=1}^n a_i^2 \ge 0.$$