Could anyone help on the following problem?
Let R(t) be the solution to the integral equation: $R(t)=1+\int_{0}^{t}\frac{1}{R(s)}ds$, namely $R(t)=\sqrt{2t+1}$. Assume that X is continuous and positive on$[0,\infty)$ and satisfies: $X(t) \leq 1+\int_{0}^{t}\frac{1}{X(s)}ds$ for $t\geq0$. Does $X(t) \leq R(t)$ follow? Either prove it or give a conterexample.
Thank you so much!
If $X(t)\ge R(t)$ for all $t$, then I think you can easily prove that $X(t)=R(t)$. So if there is a counterexample, you have to look for one which has $X(t)<R(t)$ at least part of the time.