jmc

Let $x=\frac{a+b}{a-b}$. Then, $\frac{a-b}{a+b}=\frac{1}{x}$, so the given expression is equal to $x+\frac{1}{x}=\frac{x^2+1}{x}$. Suppose that the equation $\frac{x^2+1}{x}=k$ has no solution for some value of $k$. Re-arranging, $x^2-kx+1=0$. This is a quadratic equation with discriminant $k^2-4$; since the quadratic equation has no solution, it follows that $k^2-4=(k-2)(k+2)<0$. It follows that for $k<2$, the given equation has no solution in $x$.

Thus, the smallest possible value of the given expression is $\frac{x^2+1}{x}=\boxed{2}$. Indeed, this is achievable if we take $a=1,b=0$.