The trouble with proving an inequality is that sometimes there is a very straightforward way to go about it, but for some reason it eludes you. Having gone through the audio frames for Section 3.3 of AA1 I had my head full of mathematical induction and so when asked to prove that 2n3≥(n+1)3 for n≥4 (Exercise 3.4(a)) I launched into this method without thinking. After a couple of pages of nicely worked mathematical induction I got there in the end, but the solution can actually be reached in a couple of lines.
I even think the solution in the back of the book is slightly long winded. So here is my short version:-
2n3≥(n+1)3
⇔ (3√2)n≥n+1 (Rule 5, n>0, p=3)
⇔ ((3√2)-1)n≥1 (Rule 1)
⇔ n≥1/((3√2)-1) (Rule 4, (3√2)-1>0)
1/((3√2)-1)≅3.8
Hence for integer n, 2n3≥(n+1)3 for n≥4.
The mathematical induction was at least good practice, but I wouldn't want to end up doing this in an exam. Lesson learnt!
I think I am going to enjoy analysis. It seems like we are back working with numbers and it is my sort of topic. Not that I haven't enjoyed the group theory and linear algebra - I did, but I feel like I am getting back to familiar territory.
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