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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!

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    @Jeff Tha$n$k you! I constructed one counterexample using on your hint.2012-02-14

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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) at least part of the time.