Are the sum and/or product of two increasing functions also increasing? Question:
Let $f(x)$ and $g(x)$ be two increasing functions.
a) Show that their sum is also increasing.
b) Investigate the corresponding claim for the product of two increasing functions.
Attempted solution:
a)
If $f(x)$ and $g(x)$ are increasing functions, that means that:
If $x_{1} < x_{2}$, then 
$f(x_{1}) < f(x_{2})$
$g(x_{1}) < g(x_{2})$
Adding the two inequalities gives:
$f(x_{1}) + g(x_{1}) < f(x_{2}) + g(x_{2})$
This shows that the sum is also increasing.
b)
Multiplying the original inequalities gives:
$f(x_{1})g(x_{1}) < f(x_{2})g(x_{2})$
This seems false. Even if both are increasing, one of them can be negative, and so the inequality cannot hold for all functions f and g?
Is there a way to formalize this rough intuition-based argument? Or is a counterexample the most obvious way forward? However, a counterexample seems to miss out on the more general argument about why the inequality does not hold for negative functions.
What are some productive ways to finish this question off?
 A: What can be said in general$\,$  is this:
If $f$ and $g$ are two increasing functions on an interval $I$, then:


*

*If $f(x),\, g(x)>0$ on $I$, $fg$ is increasing.

*If $f(x),\, g(x)<0$ on $I$, $fg$ is decreasing.

A: How about two negative increasing functions, that multiply to give a positive decreasing function?
$$ f(x) = g(x) = -\frac{1}{x}. $$
Then for $x>0$, $f,g$ are increasing, but $f(x) g(x) = \frac{1}{x^2}$ is decreasing.
The underlying reason is that multiplying both sides of an inequality by a negative number changes the sense of the inequality, so you get the opposite inequality if one of the functions is negative. In general, you can only multiply inequalities if both sides are known to be positive.
A: Your solution for (a) is fine, but you are right to be suspicious of (b).  The key is that you can't multiply inequalities the same way that you can add them:  In particular, from $a<b$ and $c<d$ you can only conclude $ac<bd$ if you know that all four quantities are positive.  Otherwise you could deduce obviously false things like $$-3<0 \text{ and} -4<0 \implies 12<0$$.
