Find the smallest $x$ that satisfies the inequality
$$ \left\lfloor \frac xa \right\rfloor\left\lfloor \frac xb \right\rfloor \geq y $$
where $x$, $y$, $a$, $b$ are natural numbers.
I need to solve that as part of a specific algorithm and the only thing I came up with so far is trying a lot of values for $x$. I am not sure if it can be solved analytically.
I am currently trying to find a lower bound for $x$ so that I don't have to try so many values. If I substitute $a$ and $b$ with their lowest possible value of $1$ it is clear that $x \geq \left\lfloor\sqrt{y}\right\rfloor$. That is still pretty conservative though, especially for large values of $a$ and $b$.
Is it possible to have a better lower bound or even an analytical solution for $x$?
Edit:
I realized that removing the floors can only increase the value of the left-hand side. And because the value needs to be larger, this can be used to directly solve for an $x$ that can not be too large already. (That was pointed out in the comments as well). If you solve that, you get $x \geq \left\lfloor\sqrt{aby}\right\rfloor$.
This is better than my previous lower bound. I have tested it algorithmically for values from 1 to 100 and it underestimates the $x$ by about 9% on average. There is however a worst-case, if $y$ and either $a$ or $b$ are 1 and the other $a$ or $b$ is very large.