# Comparison between solutions of ODE

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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Homework? If so, please tag it accordingly. –  Harald Hanche-Olsen Feb 14 '12 at 21:39
Thanks you for adding the homework tag. However, you ought to have kept the differential-equations tag too, or perhaps the integral-equations tag would have been better. Multiple tags are good, so long as they are appropriate. –  Harald Hanche-Olsen Feb 14 '12 at 22:18
Hint: Try setting $X(t)=1$ for $t \in [0,T]$. Then ask yourself how big the right hand side of the inequality for $X(T)$ will be compared to your exact formula for $R(T)$. Then be creative with $X(t)$ for $t\geq T$. –  Jeff Feb 14 '12 at 22:44
@Jeff Thank you! I constructed one counterexample using on your hint. –  user7762 Feb 14 '12 at 23:42
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.