Question

Show that 4n + 3^n > 6n, for all integers, n > 2.

Answer 1

Induction.

Explanation:

I will prove it with induction. For n=3:

`4(3)+3^(3) = 12 + 27 = 39 > 18.`

Suppose that the inequality is true for a k>3, that is to say

`4k+3^(k)>6k`

Then,

`4(k+1)+3^(k+1) = 4k+4+3^(k)3 = 4k+4+3^(k)+3^(k)+3^(k)`

`4k+3^(k)+4+3^(k)+3^(k) > 6k +4+1+1` because
`4k+3^(k)>6k` and
`3^(k)>1` because k>0.

Then,

`4k+3^(k)+4+3^(k)+3^(k) > 6k+6`

`4k+3^(k)+4+3^(k)+3^(k) > 6(k+1)`

`4(k+1)+3^(k+1)> (6k+1)`.

In conclusion, for all integers n> 2 the inequality is true.