# Mathematical Induction with an inequality

Use mathematical induction to prove that $$1\cdot3+2\cdot4+3\cdot5+...+n(n+2)≥ \frac{1}{3}(n^3+5n)$$ for $n≥1$.

I have tried this and got what I thought was the correct answer but it doesn't work for all values of n greater than 1. Any help would be greatly appreciated.

• Why don't you post what you have tried so we can see where you may have gone wrong? – mrp Jan 8 '15 at 10:08
• I have proved that p(1) is true, and written out the formulas for p(k) and p(k+1). I now have 1/3(k^3+5k)+(k+1)(k+3)>orequalto 1/3((k+1)^3+5(k+1)) by using the assumption that p(k) is true. If this is correct (which I'm not sure of) then I eventually get down to the inequality 3k>orequalto -1 which is obviously correct for k>orequalto 1. Any thoughts? Thanks x – Alex Jan 8 '15 at 10:10

Your reasoning is correct. Assumnig $p(k)$ true, $p(k+1)$ is equivalent to $$\frac{1}{3}(k^3+5\,k)+(k+1)(k+3)\ge\frac{1}{3}((k+1)^3+5(k+1)),$$ which reduces to $$12\,k+9\ge3\,k+6.$$ This is clearly true for all $k\ge1$.
Another chance is given by just computing the LHS. $$\begin{eqnarray*} \sum_{n=1}^{N} n(n+2) &=& \sum_{n=1}^{N}\left((n+1)^2-1\right) = -N+\sum_{n=2}^{N+1}n^2 = -(N+1)+\sum_{n=1}^{N+1}n^2\\&=&-(N+1)+\frac{(N+1)(N+2)(2N+3)}{6}=\color{red}{\frac{N(N+1)(2N+7)}{6}}.\end{eqnarray*}$$ So, we just have to check that $(N+1)(2N+7)\geq 2(N^2+5)$ for any $N\geq 1$.