I know it's kind of question that the common answer is easy to see... still looking for some more official way to formulate it.
$$ min_{y,x} y \ \ s.t. \ y \ge f(x) = min_x f(x) \ | x,y\in \mathbb{R}, f: \mathbb{R} \rightarrow \mathbb{R} $$
any suggestions?
You are given a function $f:\>{\mathbb R}\to{\mathbb R}$, and form the set $$A=\bigl\{(x,y)\in{\mathbb R}^2\bigm| y\geq f(x)\bigr\}=\bigcup_{x\in{\mathbb R}}\bigl\{(x,y)\bigm|y\geq f(x)\bigr\}\ .$$ It is then claimed that $$\min\bigl\{y\bigm|(x,y)\in A\bigr\}=\min\bigl\{f(x)\bigm| x\in{\mathbb R}\bigr\}\ ,$$ whereby we tacitly assume that these minima exist. There is not much to prove: By definition of $A$ we can write $$\min\bigl\{y\bigm|(x,y)\in A\bigr\}=\min_{x\in{\mathbb R}}\left(\min_{y\geq f(x)} y\right)=\min_{x\in{\mathbb R}}f(x)\ .$$