Let $K$ an ordered field and $a,b\in K\setminus\{0\}$. Prove that $\left|\frac{a}b+\frac{b}a\right|\ge 2$.
I give my proof but I want to know if there exist a more simple proof than that:
1) Note that $ab^{-1}ba^{-1}=1$. Then we can rewrite the inequality as $|c+c^{-1}|\ge 2$.
2) As $cc^{-1}=1>0$ then by the second axiom of order for an ordered field we know that if $c>0$ then $c^{-1}>0$. Other useful facts are
$-b=(-1)b$
$(-1)^2=1$
$a(-1)=(-1)a$
$|c^{-1}|=|c|^{-1}$
Then if $c<0$ this implies that $c^{-1}<0$. Then we can rewrite the inequality as
$$|c^{-1}c(c+c^{-1})|=|c^{-1}(c^2+1)|=|c^{-1}|(c^2+1)=|c|^{-1}(c^2+1)\ge 2$$
$$c^2+1\ge|c|2\to |c|2=|c|+|c|=2|c|\to c^2-2|c|+1\ge 0$$
Now if $c<0$ we have that $|c|=-c$ then
$$(-c)^2+2c+1=c^2+2c+1=(c+1)^2\ge 0\tag{1}$$
and for $c>0$ we similarly have that
$$c^2-2c+1=(c-1)^2\ge 0\tag{2}$$
Because $(1)$ and $(2)$ are true then the proof is complete.
My question: I searched a different proof that dont use the fact $x^2+2x+1=(x+1)^2$ but I dont found something simpler but I feel that it must exist something simpler. This is the reason why I opened this question. Maybe Im wrong but... who knows!
If $|c+\frac1{c}| < 2$, then, squaring, $c^2+2+\frac1{c^2} < 4$ or $c^2-2+\frac1{c^2} < 0$ or $(c-\frac1{c})^2 < 0$.
A contradiction.
Is this simple-minded proof correct?