Is my proof for this claim correct? Prove that for every $x,y\in\mathbb{R} $ such that $x,y>0$ and that fulfill $x\cdot y=1$, the following holds: 
$$x+y=2 \quad iff \quad x=y=1$$
Steps I took:
$$\frac { xy }{ x } =\frac { 1 }{ x } \Rightarrow y=\frac { 1 }{ x } $$
$$1+\frac { 1 }{ 1 } =2$$
Is this good enough? It feels like something is missing.
Constructing my proof using input from answers below:
Assume: $x,y\in\mathbb{R}$ such that $x,y>0$ and $xy=1$
1) $x+y=2\Rightarrow x=y=1$
$y=2-x\Rightarrow x+2-x=2$; therefore, $2-x=1$
2) $x=y=1\Rightarrow x+y=2$  
$x+(2-x)=2$
$x(2-x)=1\Rightarrow 2x-x^{ 2 }=1\Rightarrow x^2-2x+1=0$
$(x-1)^2=0\Rightarrow x=1$
$y=2-x\Rightarrow y=1$
 A: For the non-trivial way, suppose $x+y=2$, then $y=2-x$ and hence $x.(2-x)=1$, so $x^2-2x+1=0$. Now you can proceed from here.
A: In order to prove bi-conditional statements, iff, we must show that "both" directions are true. That is, if we have the statement "$A$ iff $B$", then we must show that "if $A$, then $B$. AND if $B$, then $A$", only then could we conclude $A$ iff $B$.
So, we first suppose that we have found $x,y \in \mathbb{R}$ such that $x,y > 0$ and $xy=1$. Then, we must show that (1) if $x +y =2 $, then $x=y=1$ AND (2) if $x=y=1$, then $x+y=2$. 
Direction (2) is painfully obvious, but use the comments from seeker for (1). What you have done in your post is show that if $xy=1$, then one number must be the multiplicative inverse of the other. This is fine, and in fact, true. However, this does not help prove your statement.  
A: The approaches described in the answers already posted are the most natural ones. But the following may be of interest. By expanding, we can show that for any $x$ and $y$ we have
$$(x-y)^2+4xy=(x+y)^2.\tag{1}$$
If $xy=1$ and $x+y=2$, then substituting in (1) we get $(x-y)^2=0$, and therefore $x=y$. Now from $x+y=2$ we find that $x=y=1$.
So if our two given equations hold, then $x=y=1$. And it is easy to verify that $x=y=1$ does indeed satisfy the equations.
