Proof by induction; inequality $1\cdot3+2\cdot4+3\cdot5+\dots+n(n+2) \ge \frac{n^3+5n}3$ Ok so I'm kind of struggling with this:
The question is:
"Use mathematical induction to prove that
1*3 + 2*4 + 3*5 + ··· + n(n + 2) ≥ (1/3)(n^3 + 5n) for n≥1"
Okay, so P(1) is true as 1(1+2)=3 and (1/3)(1^3 + 5)=2
Assuming P(k) is true for k≥1 gives:
1*3 + 2*4 + 3*5 + ... + k(k+2) ≥ (1/3)(k^3 + 5k)
And we want to show that P(k+1) is true, that is;
1*3 + 2*4 + 3*5 + ... + (k+1)(k+3) ≥ (1/3)((k+1)^3 + 5(k+1)) (inequality 1)
This is where I'm not sure if I'm thinking along the right lines or not. Surely, by P(k) and the rules for inequalities we get:
(1*3 + 2*4 + 3*5 + ... + k(k+2)) + (k+1)(k+3) ≥ (1/3)(k^3 + 5k) + (k+1)(k+3) (inequality 2)
but the right hand side in this inequality does not equal (1/3)((k+1)^3 + 5(k+1)) which is what I am trying to show. There's obviously a flaw in my thinking somewhere so any help would be greatly appreciated.
Thanks
Edit: As far as I understand it the right hand side of inequality 1 should equal the right hand side of inequality 2 but this isn't the case when expanded

Here's the meat of the argument:
\begin{align}
1\cdot3+2\cdot 4+\cdots+(k+1)(k+3)&\geq \frac{k^3+5k}{3}+(k+1)(k+3)\tag{ind. hyp.}\\[1em]
  &= \frac{k^3+5k+3(k^2+4k+3)}{3}\tag{simplify}\\[1em]
  &= \frac{k^3+3k^2+17k+9}{3}\tag{simplify}\\[1em]
  &\geq \frac{k^3+3k^2+8k+6}{3}\tag{since $k\geq 1$}\\[1em]
  &= \frac{(k+1)^3+5(k+1)}{3}.
\end{align}
The last equality is what you want. The key is in realizing that since $k\geq 1$ you can obtain the desired expression on the right-hand side to show that
$$
1\cdot3+2\cdot 4+\cdots+(k+1)(k+3) \geq\frac{(k+1)^3+5(k+1)}{3}.
$$
