<s>[INST] 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...
 [/INST] 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 $\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}\ge 0$
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