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.
Best Answer
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?