Inequality $\left\lfloor \frac{\sqrt{N}}{16}\right\rfloor \geq \frac{\sqrt{N}}{c}$

Question

Let $N$ be a natural number. Is it possible to show that $\left\lfloor \dfrac{\sqrt{N}}{16}\right\rfloor \geq \dfrac{\sqrt{N}}{c}$ for some constant $c$ and inequality holds for all natural $N\geq 1$?

Answer 1

It's true for any negative value of $c$.

However it cannot be true for any positive value of $c$, because for $N=1$, you have $\left\lfloor \dfrac{\sqrt{N}}{16}\right\rfloor = 0 < \frac{\sqrt N}{c}$

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If you only want the inequality to hold for large enough values of $N$, then any value $c>16$ will do the trick eventually, since, if $N$ is large, so is $$\frac{\sqrt{N}}{16} - \frac{\sqrt N}c = \frac{\sqrt{N}(c-16)}{16c},$$ meaning that $\frac{\sqrt{N}}{16} - \frac{\sqrt N}c$ will eventually be bigger than $1$.