# Induction for inequality [closed]

I have to prove by induction this inequality for $$n > 10$$: $$n-2 < \frac{n^2 - n}{12}$$

I have no idea how to start proving it. I only know that, if n=11 the inequality is true. Now, if $$n:= n+1$$ (inductive thesis) , what proceeds?

## closed as off-topic by Carl Mummert, Holo, Adrian Keister, Xander Henderson, Arnaud D.Sep 26 '18 at 12:38

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• Hypothesis: $$12(k-2)<k^2-k$$ for some $k>10$. – TheSimpliFire Sep 25 '18 at 6:24
• Now use this to show that $$12([k+1]-2)<[k+1]^2-[k+1]$$ – TheSimpliFire Sep 25 '18 at 6:25

You did the base case, now for the induction step assuming true as hypotesis

$$n-2 < \frac{n^2 - n}{12}$$

we need to show that

$$(n+1)-2 < \frac{(n+1)^2 - (n+1)}{12}$$

using the hypothesis.

Because for all $$n\geq11$$ we obtain: $$\frac{n^2-n}{12}-(n-2)=\frac{n^2-13n+24}{12}=\frac{(n-11)(n-2)+2}{12}>0.$$

• Yes, i suposed that this would be a correct answer. But the main exercise says prove by induction. Thanks anyway – Franco Cabrera Sep 25 '18 at 6:37