For statement II, since $x < y$ and both are greater than $\frac{1}{3}$, the function $f(t) = \frac{\sqrt{1+t^2}}{t}$ is increasing for $t > 0$. Thus, $f(x) < f(y)$, confirming $\sqrt{1+y^2}/y < \sqrt{1+x^2}/x$. Statement I is not always correct because $x + 1/x$ does not guarantee $< y + 1/y$ due to the non-linear nature of the function.