13. Let real numbers $a, b$ satisfy $1<a<b$. If the three numbers $1, a, b$ and $1, \frac{1}{a}, \frac{1}{b}$ cannot form the sides of a triangle, then the minimum possible value of $b$ is ( ).
(A) $\frac{3+\sqrt{3}}{2}$
(B) $\frac{5}{2}$
(C) $\frac{3+\sqrt{5}}{2}$
(D) $\frac{3+\sqrt{6}}{2}$
(E) 3
13. C.

From the problem, we know
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ a + 1 \leqslant b , } \\
{ \frac { 1 } { b } + \frac { 1 } { a } \leqslant 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\frac{1}{a} \geqslant \frac{1}{b-1}, \\
\frac{1}{b} \leqslant 1-\frac{1}{a}
\end{array}\right.\right. \\
\Rightarrow \frac{1}{b} \leqslant 1-\frac{1}{b-1} \\
\Rightarrow b^{2}-3 b+1 \geqslant 0 .
\end{array}
$$

Since $b>1$, hence $b \geqslant \frac{3+\sqrt{5}}{2}$.