Q:

Sum of strictly increasing functions is strictly increasing

Problem:
Solve
$$
\sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4} = 0
$$
Solution:
The function 
$$
\sqrt[3]{x+2}
$$
is strictly increasing, as are the functions
$$
\sqrt[3]{x+3}
$$
and
$$
\sqrt[3]{x+4}.
$$
Therefore the function we want to solve is also strictly increasing, i.e. the function has only one zero. Now just algebraically solve the equation (raise it to the power of three etc.) and get x = -3

Now, my question is, what does the summing of functions actually mean? If you compare the graph of
  $$
\sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4},
$$
  to the graphs of
  $$
\sqrt[3]{x+2}, \sqrt[3]{x+3}, \sqrt[3]{x+4},
$$
  they all go in the same direction and have exactly one zero and have some curviness, but they are still quite different. 
What happens when we add functions together? And also, it is intuitive, but why does the fact that the function's summed parts are strictly increasing make the whole function strictly increasing?

A:

Let the three strictly increasing functions be $f(x)$, $g(x)$, and $h(x)$.
Since they are strictly increasing, for $x<y$ we have the inequalities
$$f(x)<f(y)$$
$$g(x)<g(y)$$
$$h(x)<h(y)$$
Summing these inequalities gives
$$f(x)+g(x)+h(x)<f(y)+g(y)+h(y)$$
whenever $x<y$. Hence $f(x)+g(x)+h(x)$ is also a strictly increasing function.
I'm sure you can see how to generalize this.

