1. **State the problem:** We need to find the smallest value of the expression $$n + \frac{100}{n}$$ where $n$ is a positive real number. 2. **Rewrite the expression:** The function to minimize is $$f(n) = n + \frac{100}{n}$$. 3. **Find the derivative:** To find critical points, compute $$f'(n) = 1 - \frac{100}{n^2}$$. 4. **Set the derivative equal to zero:** Solve $$1 - \frac{100}{n^2} = 0$$ which simplifies to $$ \frac{100}{n^2} = 1$$. 5. **Solve for $n$:** Multiply both sides by $n^2$: $$100 = n^2$$, so $$n = \sqrt{100} = 10$$ (taking the positive root since $n > 0$). 6. **Second derivative test:** Calculate $$f''(n) = \frac{200}{n^3}$$. For $n=10$, $$f''(10) = \frac{200}{10^3} = \frac{200}{1000} = 0.2 > 0$$, indicating a local minimum. 7. **Find the minimum value:** Substitute $n = 10$ into the original function: $$f(10) = 10 + \frac{100}{10} = 10 + 10 = 20$$. **Final answer:** The smallest value of the expression is **20** at $$n = 10$$.