1. The problem involves understanding and solving inequalities and expressions involving variables $x$, $v$, constants, and exponential functions like $e^x$. 2. Let's analyze the first inequality: $x \geq e^{-x}$. 3. This inequality means we want to find all $x$ such that $x$ is greater than or equal to $e^{-x}$. 4. To solve $x \geq e^{-x}$, consider the function $f(x) = x - e^{-x}$ and find where $f(x) \geq 0$. 5. Note that $e^{-x}$ is always positive, and $x$ can be positive or negative. 6. Check the behavior at $x=0$: $f(0) = 0 - e^{0} = -1 < 0$. 7. Check at $x=1$: $f(1) = 1 - e^{-1} \approx 1 - 0.3679 = 0.6321 > 0$. 8. Since $f(0) < 0$ and $f(1) > 0$, by the Intermediate Value Theorem, there is a root between 0 and 1. 9. The root $r$ satisfies $r = e^{-r}$. 10. This root is approximately $r \approx 0.5671$. 11. For $x \geq r$, $f(x) \geq 0$, so $x \geq e^{-x}$ holds. 12. Therefore, the solution set for $x \geq e^{-x}$ is $[r, \infty)$ where $r \approx 0.5671$. Final answer: The inequality $x \geq e^{-x}$ holds for all $x \geq 0.5671$ approximately.