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.

You have to solve this inequality system:
$$\left\{\begin{matrix} 
 & \\k \neq 1 \: \land \: k\neq0
 & \\k(k+1)\geqslant0 
 & \\1+\frac{2}{k}+\frac{3}{4k^2}\geq 0
 & \\ \frac{\sqrt{k(k+1)}}{k-1} \geq 0
 & \\\left (\frac{\sqrt{k(k+1)}}{k-1}\right )^2\leq \left ( 1+\frac{2}{k}+\frac{3}{4k^2}
 \right )^2
\end{matrix}\right.$$
In this way you wil get the solutions of the inequality.
