When is it true that $x^2 < \lfloor{x}\rfloor \lceil{x}\rceil$? It seems like this should be true whenever $x$ is close to $\lfloor{x}\rfloor$ than $\lceil{x}\rceil$, but I'm not sure how to prove this. I am trying to show that this is equivalent to $\frac{x - \lfloor{x}\rfloor}{1 - (x - \lfloor{x}\rfloor)} < 1$, but I am having trouble. If someone could give me a hint about how to proceed it would be much appreciated.

Edit: Write $r = x - \lfloor{x}\rfloor$ so that $\lfloor{x}\rfloor = x - r$ and $\lceil{x}\rceil = x + (1 - r)$. Then using AM-GM, we have that

$$\frac{1}{4}((x - r) + (x + 1 - r))^2 \leq (q - r)(q + 1 - r)$$ which implies that $$\frac{1}{4}\left(2x + (1 - 2r) \right)^2 \leq \lfloor{x}\rfloor \lceil{x}\rceil$$

and it's easy to see that if $r < \frac{1}{2}$ then the LHS is larger than $x^2$. My proof does not work in the other direction though.

  • $\begingroup$ How did you get $\frac{x - \lfloor x \rfloor}{1 - (x - \lfloor x \rfloor)} < 1$? It is not equivalent to $x^2 < \lfloor x \rfloor \lceil x \rceil$, consider i.e. $x = 3/2$, or, even better $x = 1.45$. $\endgroup$
    – Ramanujan
    Apr 10, 2020 at 1:54

1 Answer 1



First notice, that when $x$ is an integer, the inequality does not hold.

Let's write $x=n+\alpha$, where $n$ is an integer and $0 < \alpha < 1$, then we can rewrite the inequality as $(n+\alpha)^2 < n(n+1)$. Now the problem is reduced to solving the following inequality:

$$\alpha^2 + 2n \alpha - n < 0$$

Can you take it from here?

  • 1
    $\begingroup$ I'm glad I refreshed before hitting enter! I had the same work. $+1$ $\endgroup$ Apr 10, 2020 at 1:56
  • $\begingroup$ @CameronWilliams So this comes out to $\alpha = \sqrt{n(n+1)} - n$. How can I understand this solution qualitatively? It looks like it gets closer to $\frac{1}{2}$ as $n$ grows large. $\endgroup$ Apr 10, 2020 at 2:23
  • $\begingroup$ @yonatano yes, this is one of the roots, the other is negative, so it is of no interest for us. The means, that the inequality holds for $\alpha \in (0, \sqrt{n(n+1)}-n)$ $\endgroup$
    – Andronicus
    Apr 10, 2020 at 2:52

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