Alternate method to prove this series in a better way Prove that $\frac{1.2 + 2.3 + 3.4 + .....+ n(n + 1)}{n(n + 3)} \ge \frac{n + 1}{4}$ for $n\ge1$
My attempt : 
 Breaking the series into two different series 
$$ S_1 = \sum_{i = 0}^n i^2 = \frac{n(n + 1) (2n + 1)}{6}$$
$$ S_2 = \sum_{i = 0}^n i = \frac{n(n + 1)}{2}$$
Now $$S_1 + S_2 = \frac{n(n + 1)(n + 2)}{3}$$
Writing down LHS and RHS we get :
$$\frac{(n + 1)(n + 2)}{3(n + 3)} \ge \frac{(n + 1)}{4}$$
Multiplying both sides by $$\frac{12}{n + 1}$$ we get : 
$$4(n + 1) \ge 3$$ 
This will always be true $$. ^ . _ . n\ge1$$
Hence proved.
Is there a mistake in my method? Is there a better method to prove this?

Since $\left(\frac{n+2}{n+3}\right)$ increases, $\left(\frac{n+2}{n+3}\right)$ has the minimum $\frac{3}{4}$ when $n=1$. Thus we get
\begin{align}
\frac{n(n+1)(n+2)}{3n(n+3)} &= \frac{(n+1)(n+2)}{3(n+3)}\\
&= \frac{n+1}{3}\cdot \frac{n+2}{n+3}\\
&\ge \frac{n+1}{3}\cdot\frac{3}{4}\\
&=\frac{n+1}{4}.
\end{align}
