I'm trying to prove the following inequality:
Let $f$ and $g$ be bounded real-valued functions with the same domain. Prove the following:
$$\inf(f) + \inf(g) \leqslant \inf(f+g).$$
I thought I had proved it, but I made the erroneous assumption that $\inf(f+g)$ can always be expressed in the form $(f+g)(x_1)$ for some $x_1$, which is not necessarily true.
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,
$\inf (f+a)= \inf (f) + a =$
$\inf (f)+ \inf(g) \le \inf (f+g)$.
The other answers already proved that $\inf(f + g) \ge \inf(f) + \inf(g)$, but for those who are curious to see an actual example of how $\inf(f + g) \gt \inf(f) + \inf(g)$ within some interval, I thought about the following one:
Let's consider the interval $[0,1]$. Let $f(x) = x + 1$ and $g(x)=\begin{cases}1 & x = 0\\ 0 & x \neq 0 \\ \end{cases}$.
Then in this interval we see that $\inf(f) = 1$, $\inf(g) = 0$ and therefore $\inf(f) + \inf(g) = 1$. Nevertheless, $f(x) + g(x) = 2$ when $x = 0$ and $f(x) + g(x) > 1$ otherwise (for instance, if $x = 0.001$, then $f(x) + g(x) = 1.001)$, and therefore $\inf(f + g) \gt \inf(f) + \inf(g)$ for all $x \in [0, 1]$.
