How can we continue at the induction step? I want to show by induction that $$\sum_{i=1}^n i^2\geq 3n+2$$ for all $n\in \mathbb{N}$ with $n\geq 3$. 
I have done the following: 
Base Case: For $n=3$ we have that $\displaystyle{\sum_{i=1}^3i^2=1^2+2^2+3^2=1+2+9=12}$ and  $\displaystyle{3n+2=3\cdot 3+2=9+2=11}$. It holds that $12\geq 11$ uand so the given inequality holds. 
Inductive hypothesis: We assume that it holds for $n=k$, so $\displaystyle{\sum_{i=1}^ki^2\geq 3k+2}$. (IH) 
Induction Step: We want to show that the inequality holds for $n=k+1$, so $\displaystyle{\sum_{i=1}^{k+1}i^2\geq 3(k+1)+2}$. 
We have that \begin{equation*}\sum_{i=1}^{k+1}i^2=\sum_{i=1}^{k}i^2+(k+1)^2\overset{(IH)}{\geq}3k+2+(k+1)^2=3k+2+k^2+2k+1=k^2+5k+3 \end{equation*} How can we continue? Can we just say that this is greater than $3k+5=3(k+1)+2$ ? 

We just need to show that
$$\sum_{i=1}^{k+1}i^2=\sum_{i=1}^{k}i^2+(k+1)^2\overset{(IH)}{\geq}3k+2+(k+1)^2\overset{(?)}\ge 3(k+1)+2$$
which is true indeed
$$3k+2+(k+1)^2\ge 3(k+1)+2$$
$$3k+2+(k+1)^2\ge 3k+5$$
$$(k+1)^2\ge 3$$
for $k\ge 2$.
