Induction/3^n \geq n^4/Exercise/Solution

Initial step for $n=10$. We have

$3^{10}=9^5=81\cdot 81\cdot 9\ge 6000\cdot 9\ge 10000=n^4.$

For the induction step let $n\ge 10$. Then

$3^{n+1}=3\cdot 3^n\ge 3\cdot n^4=n^4+\frac{1}{2}{n}^4+\frac{1}{2}{n}^4+\frac{1}{2}{n}^4+\frac{1}{2}{n}^4.$

On the other hand the general binomial formula yields

$(n+1{)}^4=n^4+4n^3+6n^2+4n+1.$

We have to show

$n^4+\frac{1}{2}{n}^4+\frac{1}{2}{n}^4+\frac{1}{2}{n}^4+\frac{1}{2}{n}^4\ge n^4+4n^3+6n^2+4n+1.$

The first summands coincide on the left and on the right. For the other summands we show that the term on the left hand side, namely $\frac{1}{2}{n}^4$, is at least as large as the term on the right hand side. But this follows directly from $n^4\ge 8n^3$ (since $n\ge 10$), from $n^4\ge 12n^2$, since $n^2\ge 12$, from $n^4\ge 8n$ and from $n^4\ge 2$.

To the exercise