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.

$$\dfrac{\sqrt{k(k+1)} }{k-1} \leq1 + \dfrac{2}{k} + \dfrac{3}{4k^2}=\dfrac{(2k+1)(2k+3)}{4k^2}$$
$$\sqrt{k(k+1)}\le\dfrac{(2k+1)(2k+3)(k-1)}{4k^2}=\frac{2k+1}{2}\times\dfrac{(2k+3)(k-1)}{2k^2}$$ We know that $\sqrt{k(k+1)}\le\dfrac{k+(k+1)}{2}$ (AM–GM inequality) and $\dfrac{(2k+3)(k-1)}{2k^2}=\dfrac{2k^2+k-3}{2k^2}$ is clearly greater than $1$ (because $k\ge3$). 
This proves the inequality.
