Proving $\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2}$ for $k \geq 3$. Could you please give me a hint on how to prove the inequality below for $k \geq 3$?

$$\frac{\sqrt{k(k+1)} }{k-1} \leq1 + \frac{2}{k} + \frac{3}{4k^2} $$ 

Thank you in advance.

If $k \geq 3, $ then...
$$\begin{align} \frac{\sqrt{k(k+1)} }{k-1} &=\frac{\sqrt{k^2+k}}{k-1} \\
&\lt \frac{\sqrt{k^2+k+\frac 1 4}}{k-1}= \frac{k+ \frac 1 2}{k-1}=1+\frac{3}{2k-2} \\
\end{align}$$
Have to compare $\frac{3}{2k-2}$ with $\frac{8k+3}{4k^2}$ when $ \ \ k \geq 3$
$$\begin{align}\frac{3}{2k-2}&=\frac{12k^2}{4k^2\cdot (2k-2)} \\
\frac{8k+3}{4k^2}&=\frac{(8k+3)(2k-2)}{4k^2 \cdot (2k-2)}=\frac{16k^2-10k-6}{4k^2 \cdot (2k-2)} \\
&=\frac{12k^2+4(k-3)^2+14(k-3)}{4k^2 \cdot (2k-2)}\end{align}$$
therefore when $ \ \ k \geq 3$
$$\frac{3}{2k-2} \leq \frac{8k+3}{4k^2}$$
And
$$\begin{align} \frac{\sqrt{k(k+1)} }{k-1} & \lt 1+\frac{3}{2k-2} \\
&=1+ \frac{12k^2}{4k^2\cdot (2k-2)}\\
&\lt 1+\frac{16k^2-10k-6}{4k^2 \cdot (k-1)} \\
&=1+\frac{8k+3}{4k^2}=1+\frac{2}{k}+\frac{3}{4k^2}
\end{align}$$
