{nsqrt(19)} = nsqrt(19) - floor(nsqrt(19)). So then we have: 19(n^2) - nsqrt(19)floor(nsqrt(19)) > 1. Now notice 4 < sqrt(19) < 5. So nsqrt(19) is between 4n and 5n. Since floor(nsqrt(19)) is an integer, it has to be 4n. So then 19(n^2) - nsqrt(19)floor(nsqrt(19)) > 19(n^2) - 16(n^2) = 3(n^2) > 1 trivially. So then we should be done.

ilikemath247365 wrote: So nsqrt(19) is between 4n and 5n. Since floor(nsqrt(19)) is an integer, it has to be 4n. this is not true

Let $[n \sqrt{19}] =a$ As $x^2 \equiv 0,1,4,9,16,6,17,11,7,5 \pmod {19}$ then $a^2 \leq 19n^2-2$ $\{n\sqrt{19}\} =n\sqrt{19}-a=\frac{19n^2-a^2}{n\sqrt{19}+a} \geq \frac{2}{n\sqrt{19}+a}>\frac{1}{n\sqrt{19}}$