How to prove that $x+y \ge 2$ for $xy = 1$ and both positive? For every $x,y\in\mathbb{R}$ such that $x,y>0$ and that satisfy the condition xy=1, the following will hold:
$$x+y\ge 2$$
Steps I took and my thoughts on this:
$$xy=1\Rightarrow \frac { xy }{ x } =\frac { 1 }{ x } \Rightarrow y=\frac { 1 }{ x } \quad and\quad xy=1\Rightarrow \frac { xy }{ y } =\frac { 1 }{ y } \Rightarrow x=\frac { 1 }{ y } $$
$$x\cdot \frac { 1 }{ x } =1\quad and\quad y\cdot \frac { 1 }{ y } =1$$
$$x+\frac { 1 }{ x } \ge 2\quad and\quad \frac { 1 }{ y } +y\ge 2$$
$$x(x+\frac { 1 }{ x } )\ge (2)x\quad and\quad y(\frac { 1 }{ y } +y)\ge (2)y$$
$$x^{ 2 }+1\ge 2x\quad and\quad y^2+1\ge 2y$$
$$x^{ 2 }-2x+1\ge 0\quad and\quad y^{ 2 }-2y+1\ge 0$$
$$(x-1)^{ 2 }\ge 0\quad and\quad (y-1)^{ 2 }\ge 0$$
$$x\in\mathbb{R} \quad and \quad y\in\mathbb{R} $$
I'm not sure where I am going wrong, but I don't know how to organize my thoughts about this in order to prove it with a formal mathematical proof. I'd like to be guided in the right direction. Hints are much appreciated.

You actually have most of the pieces; you need to stop manipulating and think about putting them together.
If $xy=1$, then $y=\frac1x$, and 
$$x+y=x+\frac1x=\frac{x^2+1}x\;.$$
Suppose that this is less than $2$; then since $x>0$, we can multiply by $x$ to find that $x^2+1<2x$, or $x^2-2x+1<0$. Now factor the lefthand side of that last inequality to get a contradiction.
