<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] 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.
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