# Infimum of functions inequality

I'm trying to prove the following inequality:

$$Let f and g be bounded real-valued functions with the same domain. Prove the following:

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

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.

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

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Wonderful! Thank you. – user64219 Oct 8 '12 at 18:54
By the way, I presume that on the first line of your proof, y > inf f + g is referring to inf(f+g)? Just clearing up any ambiguity with brackets. – user64219 Oct 8 '12 at 19:03
What else? – Did Oct 8 '12 at 19:13