Proof of $n\ge 6\implies 2n−8\le n^2−8n+ 16$ using induction.

Question

I have to do a proof by induction for this theorem:

For each $n\in \mathbb N$ such that $n\ge 6$ we have $$2n−8\le n^2−8n+ 16$$

Is this possible, or should I do a different type of proof. I am confused how to prove that $1\in S$ when $n\ge 6$, so I am thinking maybe I need to first prove $6\in S$ and then handle $n>6 $?

Thank you!

Answer 1

Base case: Let $n=6$. Then the LHS is $4$ whereas the RHS is $(6-4)(6-4)=4$, so $2n-8\le n^2-8n+16=(n-4)^2$ is true for $n=6$.

Induction Hypothesis: Assume that for some fixed $k\ge 6$ we have $$2k-8\le (k-4)^2.\tag{$I$}$$

When $n=k+1$: Suppose $n=k+1$. Then

$$\begin{align} 2(k+1)-8&=2k-6\\ &=(2k-8)+2 \\ &\le (k-4)^2+2\quad \text{(by }(I)\text{)}\\ &=k^2-8k+16+2\\ &=k^2-6k+9- (2k-9)\\ &\le k^2-6k+9\quad \text{(for }k\ge 6\text{)}\\ &=((k+1)-4)^2\\ &=(k+1)^2-8(k+1)+16, \end{align}$$

which is the same as $(I)$ but with $k$ replaced by $k+1$, so if it's true for $k$, it is true for $k+1$.

Conclusion: By induction on $n$ for all integers $n\ge 6$, we have $$2n-8\le n^2-8n+16.$$

Answer 2

The answer given by Shaun is excellent and works in general but notice here that you can factor both sides of $$2n-8\le n^2-8n+16$$

$$2(n-4)\le (n-4)(n-4)$$

and since $$n\ge 6$$ you could simplify this inequality. Do you see how to proceed ?