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I am trying to prove $n^2>2n+1$ for $k\ge 4$.

Intuitively this is true since $\lim\limits_{n\rightarrow\infty}(2+1/n)=2$.

Obviously $16>9$.

Assume $k^2>2k+1 \implies k^2+2k+2>2k+1+2k+2 \implies (k+1)^2+1>2(2k+1)+1$,

and $2(2k+1)+1>2(k+1)+1$ so $(k+1)^2+1>2(k+1)+1$, but now I am stuck.

I am not sure how to show this by induction...

Burgundy
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4 Answers4

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Assume $k^2>2k+1$; then $$ (k+1)^2=k^2+2k+1>2k+1+2k+1=4k+2=2(k+1)+1+(2k-1) $$ Can you say that $2k-1>0$?

Of course, you don't need induction at all, because this is a quadratic polynomial and we know that $$ x^2-2x-1>0 $$ for $x>1+\sqrt{2}$ or $x<1-\sqrt{2}$. In particular, $n^2-2n-1>0$ for integer $n>2$.

egreg
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Hint : $n^2-2n-1=(n-1)^2-2$ hence $n^2>2n+1$ iff $(n-1)^2>2$

Hippalectryon
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$n^2-1=(n+1)(n-1)$

$2n+1-1=2n$

Then $n+1>n$ and $n-1>2$ for $n\geq 4$.

James Pak
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$n^{2}\gt {2n+1}$ holds for $n=4$ ;

If possible let us assume that this holds upto $n=k$ for $k\ge 4$.

Now we see $$(k+1)^{2}=(k^{2})+(2k+1) \gt (2k+1)+(2k+1)=4k+2$$

Now if we can show that $$4k+2\gt 2k+3=2(k+1)+1$$ then we are done.

But $$4k+2\gt 2k+3$$ iff $$2k+2\gt 3$$ iff $$2k\gt 1$$ which is obviously true as $$k\ge 4$$ .

Thus we see $$4k+2\gt 2k+3$$ and thus proved $$(k+1)^{2}\gt 2(k+1)+1$$ So $$n^{2}\gt 2n+1$$ holds good for all $n\ge 4$.

user118494
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