Induction principle: $n^2-7n+12≥0$ for every $n≥3$

Question

How can I prove that $n^2-7n+12≥0$ for every $n≥3$?

I know that for $n=3$ I have $0≥0$ so the inductive Hypothesis is true.

Now for $n+1$ I have $(n+1)^2-7(n+1)+12=n^2-5n+6$ and now I don't know how to go on...

Answer 1

HINT

$(n+1)^2-7(n+1)+12=n^2-5n+6= (n^2 -7n+12)+(2n-6)$

Answer 2

Note that

$$n^2-5n+6=n^2-7n+12+2n-6 \stackrel{\color{red}{n^2-7n+12\ge 0}}\ge 0 + 2n-6\ge 0$$

for $n\ge3$.

Answer 3

An easy proof is to use $$n^2-7n+12=(n-3)(n-4)$$ and note that the product of positive factors is positive, and if one of the factors is zero so is the product.

No induction, no calculus, no fractions.

Answer 4

Use completing the square:

$$n^2-7n+12=\bigg(n-\frac{7}{2}\bigg)^2-\bigg(\frac{7}{2}\bigg)^2+12$$

$$\to\bigg(n-\frac{7}{2}\bigg)^2-\frac{49}{4}+\frac{48}{4}=\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}$$

Note that $k^2 \ge 0$ $\forall k \in \Bbb R$, and for $n\ge 3, \bigg(n-\frac{7}{2}\bigg)^2\ge\frac{1}{4}$, hence $\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}\ge 0$

Answer 5

Let $f(n)=n^2-7n+12$

Substitute $n=3$ and $n=4$ to get $f(n)=0$ which proves the base case true for any induction.

A method from standard high school maths is to use calculus:

Since $\frac{df}{dn}=2n-7$, the gradient is positive for $n>3.5$ so you have $f(n+1)>f(n)$ for all $n>3.5$ which gives you your successor relation to prove the theorem by induction.

This method makes a few assumptions about the function which are realistic in this case but would cause problems if e.g. the function was not continuously differentiable.

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A perhaps more elegant successor relation is obtained by discrete methods:

$\Delta f(n)=f(n+1)-f(n)=2n-6$

This says the change in $f$ from $n$ to $n+1$ is $2n-6$

So $\Delta f(n)\geq0$ for $n\geq3$ and again by induction you have your result.

Answer 6

As others have noted, this is really a silly inequality to prove using induction. Regardless, the following outline of the core part of the inductive proof may help:

\begin{align*} (k+1)^2-7(k+1)+12 &= k^2+2k+1-7k-7+12 & \text{(expand)}\\[1em] &=(k^2-7k+12)+2k-6 & \text{(rewrite to use IH)}\\[1em] &\geq0+2k-6 & \text{(by Inductive Hypothesis)}\\[1em] &\geq0+0 & \text{(since $k\geq3$)}\\[1em] &=0. \end{align*}

Answer 7

Let $f(n)=n^2-7n+12.$ We have $f(n+1)-f(n)=2n-6.$ For every $n\geq 3$ we have $$f(n)\geq 0\implies$$ $$\implies [\; f(n+1)-f(n)=2n-6\geq 0\;\land f(n)\geq 0\;]\implies$$ $$\implies f(n+1)\geq f(n)\geq 0\implies$$ $$\implies f(n+1)\geq 0.$$ So by induction we have $f(3)\geq 0\implies \forall n\geq 3\;(f(n)\geq 0).$

And we do have $f(3)\geq 0$ because $f(3)=0.$