Monotonicity of the sum/product/max of two monotone functions Suppose two monotone functions $f$ and $g$ (both weakly increasing or both weakly decreasing) are given. How can it be shown that $f+g, f \cdot g, \max(f,g)$ is again monotone (either weakly increasing or weakly decreasing)? Is there a reference to a text book?
 A: Let $f,g$ bi monotonuous increasing functions on the domain $D$. 
Then for all $x,y \in D$ with $x \leq y$ we know that $f(x)\leq f(y)$ and $g(x) \leq g(x)$. 
If we add those inequalities, we get $f(x)+g(x) \leq f(y)+g(y)$ which is equivalent to $(f+g)(x) \leq (f+g)(y)$. 
You can use this for the other two exercises.
A: To prove the product,
Let $f, g$ be two monotonuous non decreasing functions.
You have for all $x_1, x_2 \in Dom_{(f\cdot g)}$ with $x_1 \geq x_2$ holds that:
$$f(x_1) \geq f(x_2)$$ and $$g(x_1) \geq g(x_2 )$$
If $g(x_1) \geq g(x_2)$ means that $g(x_1)$ can be written as $g(x_1) = g(x_2)+k$ with $k \in \mathbb{R^+_0}$
To prove this, lets assume that $g(x_2) \geq 0 \implies g(x_1) \geq 0$ and analogous for $f(x_2) \geq 0 \implies f(x_1) \geq 0$
So, multiplyng by $f(x_1) \geq f(x_2)$ by $g(x_2)$ you have $f(x_1)g(x_2) \geq f(x_2)g(x_2)$
Since $f(x_1) \geq 0\implies f(x_1) + k \geq 0$ ( a weak inequality )
So, add up to the LHS of the inequation $f(x_1)g(x_1) + f(x_1)+k\geq f(x_2)g(x_1)$
$$f(x_1)\left[g(x_2) + k\right] \geq f(x_2)g(x_1)$$, that is
$$f(x_1)\left[g(x_1)\right] \geq f(x_2)g(x_1) = (f\cdot g)(x_1) \geq (f\cdot g)(x_2) \iff f,g$$ are monotonuous non decreasing functions $\land f(x), g(x) \geq 0 \forall x \in Dom_{(f\cdot g)}$.
