A question on floor function. Prove that for $n \in N$,
$[\sqrt{n} + \frac{1}{2} ] = [\sqrt{n-\frac{3}{4}} + \frac{1}{2}]$
where [.] is the greatest integer function.
My attempt: k < $\sqrt{n} + \frac{1}{2} $ < k+1 .
Obviously, $\sqrt{n-\frac{3}{4}} + \frac{1}{2}$ < $\sqrt{n} + \frac{1}{2} $,
so it would suffice to show that $\sqrt{n-\frac{3}{4}} + \frac{1}{2}$> k , but it turns out that the inequlity is very weak so i am not able to prove it.
Is there any way to do this using induction? Or am i trying the right thing?
Any help is appreciated , thanks!

We have, where $k$ is the integer part of $\sqrt{n} + \frac{1}{2}$,
$$\begin{equation}\begin{aligned}
k & \lt \sqrt{n} + \frac{1}{2} \\
k - \frac{1}{2} & \lt \sqrt{n} \\
k^2 - k + \frac{1}{4} & \lt n
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Since $k$ and $n$ are positive integers, \eqref{eq1A} can actually be restated as being
$$\begin{equation}\begin{aligned}
k^2 - k + 1 & \le n \\
k^2 - k + \frac{1}{4} & \le n - \frac{3}{4} \\
\left(k - \frac{1}{2}\right)^2 & \le n - \frac{3}{4} \\
k - \frac{1}{2} & \le \sqrt{n - \frac{3}{4}} \\
k & \le \sqrt{n - \frac{3}{4}} + \frac{1}{2}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $\sqrt{n} + \frac{1}{2} \gt \sqrt{n - \frac{3}{4}} + \frac{1}{2}$, as you stated, this shows $k$ is also the integer part of $\sqrt{n - \frac{3}{4}} + \frac{1}{2}$, which means that
$$\left\lfloor \sqrt{n} + \frac{1}{2} \right\rfloor = \left\lfloor \sqrt{n - \frac{3}{4}} + \frac{1}{2} \right\rfloor \tag{3}\label{eq3A}$$
