smc

Notice that $1>\frac{1}{a}>\frac{1}{b}$. Using the triangle inequality, we find $$a+1>b⟹a>b-1$$ $$\frac{1}{a}+\frac{1}{b}>1$$ In order for us the find the lowest possible value for $b$, we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get $$a=b-1$$ and $$\frac{1}{a}+\frac{1}{b}=1$$ Substituting, we get $$\frac{1}{b-1}+\frac{1}{b}=\frac{b+b-1}{b(b-1)}=1$$ $$\frac{2b-1}{b(b-1)}=1$$ $$2b-1=b^2-b$$ Solving for $b$ using the quadratic equation, we get $$b^2-3b+1=0⟹b=\boxed{\frac{3+\sqrt{5}}{2}}$$