Struggling with questions about discrete math Induction

Question

Prove my induction that "" n^2 + 7n + 3 ≤ 3 n^2 for n ≥ 4 ""

I know that the base case is n=4 and I assume that n= k is true and I have to prove that it is also true for n= k+1

but I don't know how to go about starting that from what I have

Suggested reading: How to write a clear induction proof. – JMoravitz Oct 07 '20 at 16:38 — JMoravitz, Oct 07 '20 at 16:38

score 1 · Answer 1 · answered Oct 07 '20 at 16:37

1

We need to prove that: $$2n^2-7n-3\geq0$$

For $n=4$ we obtain $1\geq0$, which is true.

Now, by assumption of the induction for $n\geq4$ we obtain: $$2(n+1)^2-7(n+1)-3=2n^2-3n-8=2n^2-7n-3+4n-5>0.$$

Thus, by the axioma of an induction we are done.

answered Oct 07 '20 at 16:37

Michael Rozenberg

The final inequality could use a bit more detail. $\underbrace{(2n^2-7n-3)}{\geq 0~\text{by I.H.}} + \underbrace{(4n-5)}{\geq 16-5~\text{since }n\geq 4}>0$ – JMoravitz Oct 07 '20 at 16:43
@JMoravitz Thank you for your interest! – Michael Rozenberg Oct 07 '20 at 16:45
thanks for the response – Minusa Oct 07 '20 at 17:39
@Minusa You are welcome! – Michael Rozenberg Oct 07 '20 at 17:40

score 1 · Answer 2 · answered Oct 07 '20 at 16:39

1

You want to show that $2n^2\geq 7n+3$ for $n\geq 4$.

Induction start

For $n=4$ we have $2n^2=32\geq31=7n+3$

For $n=5$, $2n^2=50\geq 38=7n+3$

Induction hypothesis

Assume it holds for $n=k$ ($\geq 4$).

Induction step

$$2(k+1)^2=2(k^2+2k+1)$$ $$=2k^2+4k+2\geq7k+3+4k+2 \geq 7k+5+4\cdot4$$ $$>7k+10=7(k+1)+3$$

Thus $2(k+1)^2\geq7(k+1)+3$ as required.

answered Oct 07 '20 at 16:39

Alessio K

2 Answers2