<s>[INST] 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) $x=y=1\Rightarrow x+y=2$  
$x+(2-x)=2$
$x(2-x)=1\Rightarrow 2x-x^{ 2 }=1\Rightarrow x^2-2x+1=0$
$(x-1)^2=0\Rightarrow x=1$
$y=2-x\Rightarrow y=1$
 [/INST] 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.
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