$ \inf\{(f+g)(x)\mid x\in[a,b] \}\geq \inf\{g(x)\mid x\in[a,b] \}+ \inf\{f(x)\mid x\in[a,b] \} $

Question

Prove that $$ \inf\{(f+g)(x)\mid x\in[a,b] \}\geq \inf\{g(x)\mid x\in[a,b] \}+ \inf\{f(x)\mid x\in[a,b] \} $$ I need this for proving $\int f+g=\int f+ \int g$.
Where do I begin ? I do not have an idea why this should be true.

Answer 1

You start with $$ f(x)+g(x)\ge\inf_{s\in[a,b]} f(s)+\inf_{s\in[a,b]} g(s) $$ and proceed from there.

Answer 2

The lowest possible value of $f+g$ should be no lower than the lowest value of $f$ plus the lowest value of $g$. To see why this might be true, consider what would happen if

1. the lowest point of $f$ and the lowest point of $g$ happen for the same $x$-value
2. they happen for different $x$-values

In case 1. we clearly have equality. In case 2, let $s$ be the point that gives $f(s)+g(s)$ the smallest possible value, let $t$ be the point that gives $f(t)$ the smallest possible value, and $u$ is such that $g(u)$ is as small as possible. Now compare $f(s)+g(s)$ to $f(t)+g(u)$. One is clearly smaller than the other.

This is an intuitive explanation of why it should be true. For a full proof, you need to go into the definition of $\inf$ and deal with all the details. That is important, but in my opinion doesn't really teach you much unless you understand the intuition.