How can we continue at the induction step?

Question

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 $3k+5=3(k+1)+2$ ?

Answer 1

Must it not be $$\geq 3(k+1)+2$$? Why don't you use $$\sum_{i=1}^ni^2=\frac{1}{6} n (n+1) (2 n+1)$$

Answer 2

I don't know what is the reason to prove that even though you want to demonstrate induction, but you can finish it very simple. Since k > 2, we have:

$$ 3k+2+(k+1)^2 > 3k + 2 + (2 + 1)^2 = 3k + 2 + 9 = 3k + 11 > 3k + 5 $$

Also, correct your base, it's 14. not 12 :-)

Answer 3

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 $k\ge 2$.