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

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...

• A prood without induction would be far easier. May 10 '18 at 18:12
• yes, using for instance $g(x)=x^2-7x+12$ and working with derivatives May 10 '18 at 18:14
• @AlexA simpler than that by noting $n^2-7n+12 = (n-3)(n-4)$ and so for $n=3$ and $n=4$ the expression is zero and for any $n$ greater than that it is the product of two positive numbers. May 10 '18 at 18:18
• Without induction: $n^2-7n+12 = (n-4)^2+(n-4) \ge 0$ for $n\ge 4$. The case $n=3$ is easy.
– lhf
May 10 '18 at 19:02
• @user557276 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– user
May 31 '18 at 20:49

HINT

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

• How do I know that $2n-6$ is greater or equal to zero too? May 10 '18 at 18:57
• @user557276 If $n \ge 3$, you are good. That is one of your hypotheses. May 10 '18 at 18:58

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.

• The exercise asks to use induction May 10 '18 at 18:55
• @user557276 You used the word induction and the tags induction and calculus (calculus has now gone. The first line of your post is a question which requires neither of these techniques. I therefore felt able to suggest an alternative answer. I did not know whether you were trying to use induction to solve a general problem or not because you didn't tell me. In any event you can use induction to show that the product of positive integers is positive if you so wish. And the alternatives to induction are usually worth exploring (some of them hide the induction instead of making it explicit). May 10 '18 at 19:09

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*}

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 $n\ge3$.

• I thought it was clear since $2n-6\ge 0$ for $n\ge0$. I ad that, Thanks!
– user
May 10 '18 at 18:49
• @hardmath But that inequality is the induction hypothesis, which the OP is aware they are assuming. May 10 '18 at 19:01
• @B.Mehta Thanks for the explanation I've added something to make that point more clear.
– user
May 10 '18 at 19:04
• @hardmath I've added some more detail.
– user
May 10 '18 at 19:04
• Note that content is intended for future Readers as well. I appreciate your taking the trouble to respond, and I was not a downvoter. May 10 '18 at 19:16

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.

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.

• A downvote without a comment is an unkind act. Jun 5 '18 at 21:24

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 $f(3)=0.$

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 $\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}\ge 0$

• The question specifically asks about an induction proof. This doesn't seem to answer the question. May 10 '18 at 18:59
• How is this even remotely by far the most obvious proof? Simply note that $P(n)=n^2-7n+12=(n-3)(n-4)$, whereby we note that $P$ is negative (for real $n$) only on the interval $(3,4)$ and nonnegative otherwise. In particular, $P$ is zero for $n=3,4$ and positive for all $n>4$. Since we are only interested in the nonnegativity of $P$ for integer values of $n$, the "proof" almost immediately follows from the factorization of the monic trinomial. May 10 '18 at 20:08
• @DanielW.Farlow agreed for this case your proof is easier, but as a general case for proving non-negativity, CTS seems to be a more obvious method (at least to me) May 10 '18 at 21:50
• @XanderHenderson, this is a case where an inductive proof is both overcomplicated and unnecessary. Hence, I proved it with a method more suited to this style of problem. May 10 '18 at 21:53
• @RhysHughes I don't disagree. But the question isn't "How do I prove this?" The question is "How do I finish this proof by induction?" May 10 '18 at 22:00