May be easy to guess,but hard to prove^_^

Notation: Here $(a,b)$ stands for the greatest commom divisor of two integers . And $\lfloor x \rfloor$ stands for the greatest integer that no larger than a real number $x$ .

The answer is 8 and the following proof consists of two parts, Part1showes 8 is ok ,Part2 showes any postive real numbers less than 8 will fail the inequality for some particular $n_{\in Z^{+}}$ using Pell's Equation .(See here: 1.[https://brilliant.org/wiki/quadratic-diophantine-equations-pells-equation/) 2.[https://en.wikipedia.org/wiki/Pell%27s_equation)

Part1. $\large\forall n\in Z^{+}$ ,we have $\large(n,\lfloor n\sqrt{2} \rfloor)<8^{\frac{1}{4}}\sqrt{n}$ .

Since $(n\sqrt{2})^{2}=2n^{2}$ is square-free,there exists $a\in Z^{+}$ such that $a^{2}<2n^{2}<(a+1)^{2}$ .And now $\lfloor n\sqrt{2} \rfloor =a$ .

Let $d=(n,\lfloor n\sqrt{2} \rfloor)$ ,then $d^{2}|n^{2}, d^{2}|a^{2}$ .So $d^{2}|(2n^{2}-a^{2})_{>0}$ ,which gives us an inequality: $d^{2}\le 2n^{2}-a^{2}$ $\le ((a+1)^{2}-1)-a^{2}=2a<2\sqrt{2}n$ .Thus, $d=(n,\lfloor n\sqrt{2} \rfloor)<8^{\frac{1}{4}}\sqrt{n}$ .

Part2. $\large\forall 0<\delta <8,\exists n\in Z^{+}$ $\large:$ $\large(n,\lfloor n\sqrt{2} \rfloor)>(8-\delta)^{\frac{1}{4}}\sqrt{n}$ .

Let us consider this Negative Pell Equation: $y^{2}-2x^{2}=-1$ .It has infinitely many distinct solutions based on the obvious solution $(1,1)$ .

Now we can take one of its solutions, say $(x,y)$ , of which $x>\lfloor \sqrt{\frac{4}{\delta}} \rfloor +1$ . Then $4y^{2}+4=8x^{2}>(8-\delta)x^{2}+4$ .

Take n=2xy . Then $2y^{2}<2n^{2}=8x^{2}y^{2}=4y^{2}(y^{2}+1)<(2y^{2}+1)^{2}$ , i.e. $2y^{2}<\sqrt{2}n<2y^{2}+1$ , which means $\lfloor \sqrt{2}n \rfloor=2y^{2}$ .

Note that $(x,y)=1$ , we see $d=(n,\lfloor n\sqrt{2} \rfloor)=(2xy,2y^{2})=2y$ . Therefore, $d^{4}=16y^{4}>(8-\delta)x^{2}-4y^{2}=(8-\delta)n^{2}$ ,which implies $d=(n,\lfloor n\sqrt{2} \rfloor)>(8-\delta)^{\frac{1}{4}}\sqrt{n}$ . $\blacksquare$