## anonymous one year ago Show that n^3 > (n+1)^2 for all natural numbers n>=3

1. ganeshie8

$$n\ge 3$$ multiply both sides by $$n^2$$ and get $$n^3\ge 3n^2\\=n^2+n^2+n^2 \\=n^2 + n\cdot n+n^2 \\\gt n^2 + 2\cdot n + 1 \\= (n+1)^2$$

2. mathmate

@ganeshie8 : brilliant! Or if we prefer the long way, using mathematical induction, Statement: $$n^3>(n+1)^2$$ for n>=3. Base case: for n=3, we have $$n^3-(n+1)^2=3^3-4^2=11 > 0$$ so statement holds for n=3. Inductive Hypothesis: assuming statement holds for case n=k. i.e. $$k^3-(k+1)^2 = S_k >0$$ Inductive step: for case n=k+1, we have $$(k+1)^3-(k+1+1)^2 =k^3+3k^2+3k+1-(k^2+4k+4)$$ $$=[k^3-(k^2+2k+1)] + 3k^2+k-2)$$ $$=S_k +(3k-2)(k+1)$$ On the last line, $$S_k >0$$ by hypothesis (3k-2) > 0 $$\forall k\ge3$$ (k+1) >0 $$\forall k\ge3$$ Therefore $$=S_k +(3k-2)(k+1) > 0$$ for case n=k+1 Conclusion: By mathematical induction, we have established that statement $$n^3>(n+1)^2$$ is true for all integers n>=3

3. Jhannybean

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