$n=2k$ Let $[2k\sqrt{6}]=m$ Case 1: $m=2t$ Then $ 24k^2-4t^2=4(6k^2-t^2) \geq 8$ Case 2: $m=2t+1$ $24k^2-(2t+1)^2=c$ $24k^2-4t^2-4t=c+1 \to 8|c+1 \to c \geq 7$ If $c=7$ then $6k^2=t^2+t+2$ has no solutions, so $c \geq 15$ So $24k^2-m^2 \geq 8$ $\{2k\sqrt6\}=2k\sqrt{6}-[2k\sqrt{6}]=\frac{24k^2-m^2}{2k\sqrt{6}+m} >\frac{4}{2k\sqrt{6}}=\frac{4}{n\sqrt{6}}$ b )$ \frac{4}{n\sqrt{6}}> \frac{1}{n-\frac{1}{5n}}$ $4(5n^2-1)>5\sqrt{6}n^2$ $5(4-\sqrt{6})n^2>4$ - true for all $n$

RagvaloD wrote: $\frac{24k^2-m^2}{2k\sqrt{6}+m} >\frac{4}{2k\sqrt{6}}$ Why?

any ideas?

Well, as shown in the argument the numerator is at least $8$, so the fraction is at least $\frac{8}{2k\sqrt{6}+m}>\frac{8}{4k\sqrt{6}}=RHS$ since $m<2k\sqrt{6}$ by definition.

Claim 1: The following terms are never squares: \[24k^2, 24k^2-1,24k^2-2,24k^2-3,24k^2-4,24k^2-5,\ldots 24k^2-7\]Proof: The only quadratic residues mod 24 are 0,1,4,9,12,16, which deal with $24k^2-1,\ldots 24k^2-7$. Then, note that $24k^2$ can never be a square because $v_3(24k^2)\equiv 1\pmod{2}$. Claim 2: For positive even integers $n$, \[\sqrt{6n^2}- \sqrt{6n^2-8}>\frac{1}{n-\frac{1}{5n}}\]Proof: Note that, \[\frac{1}{n-\frac{1}{5n}}^2 + 2\cdot \frac{1}{n-\frac{1}{5n}} \cdot \sqrt{6n^2-8}+6n^2-8 < 1+ 2\cdot \frac{1}{\sqrt{n^2-\frac{8}{6}} } \sqrt{6n^2-8} +6n^2-8 \]\[=1+2\sqrt{6}+6n^2-8 < 6n^2\]$\square$ Applying Claim 1, for even $n$, \[\lfloor \sqrt{6}n \rfloor \leq \sqrt{6n^2-8}\]Then, applying Claim 2, \[\{n\sqrt{6}\}=n\sqrt{6}-\lfloor \sqrt{6}n\rfloor \geq \sqrt{6n^2}- \sqrt{6n^2-8}>\frac{1}{n-\frac{1}{5n}}\]$\blacksquare$.