Certainly! We can prove the inequality \( f(x+y) \leq f(x) + f(y) \) using calculus by leveraging the condition \( f''(x) \leq 0 \), which implies that \( f \) is concave. Here’s the proof:

### Proof:

1. **Define a new function:**
   Define a function \( g(x) \) as follows:
   \[
   g(x) = f(x + y) - f(x) - f(y),
   \]
   where \( y \) is fixed and \( x \) varies. We need to show that \( g(x) \leq 0 \) for all \( x \geq 0 \).

2. **Compute the first derivative of \( g(x) \):**
   Differentiate \( g(x) \) with respect to \( x \):
   \[
   g'(x) = \frac{d}{dx}g(x) = \frac{d}{dx} \left( f(x+y) - f(x) - f(y) \right) = f'(x+y) - f'(x).
   \]

3. **Compute the second derivative of \( g(x) \):**
   Differentiate \( g'(x) \) with respect to \( x \) to obtain the second derivative:
   \[
   g''(x) = \frac{d}{dx}g'(x) = \frac{d}{dx} \left( f'(x+y) - f'(x) \right) = f''(x+y) - f''(x).
   \]
   Given that \( f''(x) \leq 0 \) for all \( x \geq 0 \), this implies \( f''(x+y) \leq 0 \) and \( f''(x) \leq 0 \). Therefore,
   \[
   g''(x) = f''(x+y) - f''(x) \leq 0 - 0 = 0.
   \]
   Hence, \( g''(x) \leq 0 \), which implies that \( g'(x) \) is a non-increasing function.

4. **Analyze \( g'(x) \):**
   Since \( g'(x) = f'(x+y) - f'(x) \) is non-increasing, \( g'(x) \) either stays constant or decreases as \( x \) increases.

   - **Case 1: \( g'(x) \) is constant.** If \( g'(x) = 0 \) for all \( x \geq 0 \), then \( f'(x+y) = f'(x) \), which implies that \( f(x+y) - f(x) \) is constant, and thus \( g(x) = \) constant. Since \( f \) is concave, the constant value of \( g(x) \) must be non-positive, implying \( g(x) \leq 0 \).

   - **Case 2: \( g'(x) \) decreases.** If \( g'(x) \) is decreasing, then \( g'(x) \) will eventually become non-positive, meaning \( f'(x+y) \leq f'(x) \) for all \( x \geq 0 \). This implies that \( g(x) \) is non-increasing. Since \( g(x) \) is non-increasing and \( g(x) = 0 \) when \( x = 0 \), it follows that \( g(x) \leq 0 \) for all \( x \geq 0 \).

5. **Conclusion:**
   Since \( g(x) = f(x+y) - f(x) - f(y) \leq 0 \), we conclude that:
   \[
   f(x+y) \leq f(x) + f(y),
   \]
   for all \( x, y \geq 0 \).

This completes the proof using calculus that \( f(x+y) \leq f(x) + f(y) \) given that \( f''(x) \leq 0 \).

Would you like more details or have any questions?

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Here are 8 related questions that might interest you:
1. How does concavity affect the behavior of the first derivative \( f'(x) \)?
2. Can you provide an example where \( f(x+y) = f(x) + f(y) \)?
3. What would the inequality look like if \( f \) were convex instead of concave?
4. How does the proof change if we assume \( f(x) \) is defined on a different domain?
5. Can we extend this proof to functions of multiple variables?
6. How does the second derivative test relate to this proof?
7. Can this proof be generalized to functions with higher-order derivatives?
8. What are the implications of this inequality in optimization problems?

**Tip:** When analyzing functions using calculus, consider the implications of the second derivative, as it provides key insights into the function's concavity and convexity.

Explore the proof of the inequality f(x+y) <= f(x) + f(y) using calculus. Understand the implications of concavity and the second derivative test in mathematical proofs. Suitable for advanced undergraduate or graduate-level studies.

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