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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?

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

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

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

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

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