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...
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...
HINT
$(n+1)^2-7(n+1)+12=n^2-5n+6= (n^2 -7n+12)+(2n-6)$
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.
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 $n\ge3$.
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.
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 $\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}\ge 0$