Prove that the sum of the infima is smaller than the infimum of the sum

Let $h=f+g$ and $y\gt\inf h$, then there exists $x$ such that $y\geqslant h(x)=f(x)+g(x)$. But $f(x)\geqslant\inf f$ and $g(x)\geqslant\inf g$ hence $y\geqslant\inf f+\inf g$.

Every $y\gt\inf h$ is such that $y\geqslant\inf f+\inf g$. Hence $\varepsilon+\inf h\geqslant\inf f+\inf g$, for every $\varepsilon\gt0$. In particular, $\inf\{\varepsilon+\inf h\mid\varepsilon\gt0\}\geqslant\inf f+\inf g$. The infimum of the set on the LHS is $\inf h$ hence all this proves that $\inf h\geqslant\inf f+\inf g$.

Likewise, $\sup h\leqslant\sup f+\sup g$.

It seems that we can argue in the following simple way and no need to introduce $\varepsilon$: Denote $a=\inf(f)$ and $b=\inf(g)$. Note that for each $x$ in the domain, $a\leq f(x)$ and $b\leq g(x)$. Therefore, $a+b\leq f(x)+g(x).$ This shows that $a+b$ is a lower bound of $\{(f+g)(y)\mid y\in\mbox{Domain}\}$ and hence $a+b\leq\inf(f+g)$.

Another attempt:

Let $a:=\inf (g)$, then $a \le g$, and

$f+a \le f +g$.

Hence

$\inf (f +a) \le \inf (f+g).$

Since $a$ is fixed,

$\inf (f+a)= \inf (f) + a =$

$\inf (f)+ \inf(g) \le \inf (f+g)$.