# When is it true that $x^2 < \lfloor{x}\rfloor \lceil{x}\rceil$?

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.

• 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 $x = 1.45$. – Viktor Glombik Apr 10 at 1:54

Hint:

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?

• I'm glad I refreshed before hitting enter! I had the same work. $+1$ – Cameron Williams Apr 10 at 1:56
• @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. – TheProofIsTrivium Apr 10 at 2:23
• @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 $\alpha \in (0, \sqrt{n(n+1)}-n)$ – Andronicus Apr 10 at 2:52