Examples for $\max\{f(x)+g(x)\}\leq \max{f(x)}+ \max{g(x)}$ and $\min\{f(x)+g(x)\}\geq\min{f(x)}$+min{g(x)}.

Question

Can someone give me an example with concrete functions for the following relations $\max\{f(x)+g(x)\}\leq \max{f(x)}+ \max{g(x)}$ and $\min\{f(x)+g(x)\}\geq \min{f(x)}+\min{g(x)}$? I suppose one example would be enough to verify both.

Answer 1

Try the following: $f(x)=1$ and $g(x)=2$.

Another example.

$$f(x)=\frac{x}{x^2+1}$$ and $$g(x)=\frac{x}{x^2+4}$$.

$$\max{f}+\max{g}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}>\max(f+g)=0.71...$$

$$\min{f}+\min{g}=-\frac{1}{2}-\frac{1}{4}=-\frac{3}{4}<\min(f+g)=-0.71...$$

Answer 2

Take $f(x) = \sin x$ and $g(x) = \cos x$.

Answer 3

For $x\in [-\frac{\pi}{2},\frac{\pi}{2}]$ set $g(x)=\cos(x)$ and $f(x)=\sin(x)$.

Answer 4

If you mean an example of strict inequality, a pretty extreme example would be $f(x) = x$ and $g(x) = -x$. Then the max of the sum is $0$ but the sum of the maxes is unbounded. Also works for min.

Answer 5

Another solution: $f(x)=g(x)=0$ (or equal to any other constants)