Real numbers $a$ and $b$ are chosen with $1<a<b$ such that no triangle with positive area has side lengths $1,a,$ and $b$ or $\tfrac{1}{b}, \tfrac{1}{a},$ and $1$. What is the smallest possible value of $b$?

${ \textbf{(A)}\ \dfrac{3+\sqrt{3}}{2}\qquad\textbf{(B)}\ \dfrac52\qquad\textbf{(C)}\ \dfrac{3+\sqrt{5}}{2}\qquad\textbf{(D)}}\ \dfrac{3+\sqrt{6}}{2}\qquad\textbf{(E)}\ 3 $
1. **Understanding the problem**: We need to find the smallest possible value of \( b \) such that no triangle with side lengths \( 1, a, b \) or \( \frac{1}{b}, \frac{1}{a}, 1 \) has positive area. This means that the side lengths do not satisfy the triangle inequality.

2. **Triangle inequality for \( 1, a, b \)**: For a triangle with sides \( 1, a, b \) to not have positive area, it must fail the triangle inequality. Since \( b \) is the largest side, the inequality \( b \ge a + 1 \) must hold. To minimize \( b \), we set \( b = a + 1 \).

3. **Triangle inequality for \( \frac{1}{b}, \frac{1}{a}, 1 \)**: Similarly, for the triangle with sides \( \frac{1}{b}, \frac{1}{a}, 1 \) to not have positive area, it must fail the triangle inequality. Since \( \frac{1}{a} \) and \( \frac{1}{b} \) are less than 1, the inequality \( \frac{1}{b} + \frac{1}{a} \le 1 \) must hold. To minimize \( b \), we set \( \frac{1}{b} + \frac{1}{a} = 1 \).

4. **Substitute \( a = b - 1 \)**: From step 2, we have \( a = b - 1 \). Substitute this into the equation from step 3:
   \[
   \frac{1}{b} + \frac{1}{b-1} = 1
   \]

5. **Solve the equation**:
   \[
   \frac{1}{b} + \frac{1}{b-1} = 1
   \]
   Multiply both sides by \( b(b-1) \):
   \[
   (b-1) + b = b(b-1)
   \]
   Simplify:
   \[
   2b - 1 = b^2 - b
   \]
   Rearrange to form a quadratic equation:
   \[
   b^2 - 3b + 1 = 0
   \]

6. **Solve the quadratic equation**:
   \[
   b = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}
   \]
   Since \( b > 1 \), we take the larger root:
   \[
   b = \frac{3 + \sqrt{5}}{2}
   \]

Conclusion:
\[
\boxed{\frac{3 + \sqrt{5}}{2}}
\]