The problem given is written in Persian, and I will translate and solve it:

Translation:

"Assume $f$ and $g$ are increasing functions, and at least one of them is strictly increasing. Prove that: $f(x) \leq g(x) \quad \forall x \in \mathbb{R}$ implies $f(f(x)) \leq g(g(x)) \quad \forall x \in \mathbb{R}.$"

Solution Outline:

1. $f$ and $g$ are increasing functions, which means if $x_1 \leq x_2$, then: $f(x_1) \leq f(x_2) \quad \text{and} \quad g(x_1) \leq g(x_2).$

2. We are given $f(x) \leq g(x)$ for all $x \in \mathbb{R}$.

3. Since $f(x) \leq g(x)$, applying $f$ and $g$ to both sides preserves the inequality, because both functions are increasing. Thus: $f(f(x)) \leq f(g(x)).$

4. Similarly, since $f(x) \leq g(x)$, and applying $g$ to both sides (again due to the increasing nature of $g$): $g(f(x)) \leq g(g(x)).$

5. We need to show that $f(f(x)) \leq g(g(x))$, which follows directly from the above steps since $f(f(x)) \leq f(g(x)) \leq g(g(x))$.

Thus, the desired inequality is proved.

Would you like more details on any step of this proof?

Here are some related questions for further practice:

1. What changes if both $f$ and $g$ are strictly increasing?
2. Can you provide a counterexample if neither $f$ nor $g$ are increasing?
3. How does the proof change if $f$ and $g$ are only non-decreasing instead of strictly increasing?
4. Can you extend this result to compositions of more than two functions?
5. What if $f$ and $g$ were decreasing functions instead?

Tip: When dealing with increasing functions, inequalities are preserved when applying the function to both sides.