Proof by induction; inequality $1\cdot3+2\cdot4+3\cdot5+\dots+n(n+2) \ge \frac{n^3+5n}3$

Question

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

Answer 1

Assume that statement is true for number unto k. We need to show that it is true for k+1.

$$ 1*3+...+k(k+2)+(k+1)(k+3)\geq \frac{k^3+5k}{3}+(k+1)(k+3)=\frac{k^3+3k^2+17k+9}{3}\geq\frac{k^3+3k^2+8k+6}{3}=\frac{(k+1)^3+5(k+1)}{3} $$

for last inequality expand both sides and see that

$$ \frac{(k+1)^3+5(k+1)}{3}=\frac{k^3+3k^2+8k+6}{3} $$

and $$ \frac{k^3+5k}{3}+(k+1)(k+3)=\frac{k^3+3k^2+17k+9}{3} $$

put it in original inequation and get the result

Answer 2

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 $$ 1\cdot3+2\cdot 4+\cdots+(k+1)(k+3) \geq\frac{(k+1)^3+5(k+1)}{3}. $$