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.
 A: We have
$$\left(1+\frac{2}{k}+\frac{3}{4k^2}\right)^2-\frac{k(k+1)}{(k-1)^2}=1/16\,{\frac {16\,{k}^{5}-24\,{k}^{4}-64\,{k}^{3}+{k}^{2}+30\,k+9}{{k}
^{4} \left( k-1 \right) ^{2}}}
\geq 0$$ for $k\geq 3$
A: 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
$$\begin{align} \frac{\sqrt{k(k+1)} }{k-1} & \lt 1+\frac{3}{2k-2} \\
&=1+ \frac{12k^2}{4k^2\cdot (2k-2)}\\
&\lt 1+\frac{16k^2-10k-6}{4k^2 \cdot (k-1)} \\
&=1+\frac{8k+3}{4k^2}=1+\frac{2}{k}+\frac{3}{4k^2}
\end{align}$$
A: $$\dfrac{\sqrt{k(k+1)} }{k-1} \leq1 + \dfrac{2}{k} + \dfrac{3}{4k^2}=\dfrac{(2k+1)(2k+3)}{4k^2}$$
$$\sqrt{k(k+1)}\le\dfrac{(2k+1)(2k+3)(k-1)}{4k^2}=\frac{2k+1}{2}\times\dfrac{(2k+3)(k-1)}{2k^2}$$ 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.
A: 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.
