Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.