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I'm currently trying to prove the inequality $$n^2<n(n+2)<(n+1)^2$$

Is it possible to solve this without induction?

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    $\begingroup$ Just expand them. $\endgroup$ – Theo Bendit May 22 at 15:35
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    $\begingroup$ So I got $n^2<n^2 +2n<n^2 +2n+1$ and $0<2n<2n+1$ by subtracting $n^2$ is this correct? $\endgroup$ – Loo Soo Yong May 22 at 15:37
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    $\begingroup$ Yeah, that's right. Since $n$ is positive, $0 < n$, so $0 < 2n$. And, always, $0 < 1$ so $2n < 2n + 1$. Add $n^2$ to all sides, and you have your inequality. $\endgroup$ – Theo Bendit May 22 at 15:39
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Expanding we get $$n^2<n^2+2n<n^2+2n+1$$ Can you proceed?

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  • $\begingroup$ Can I subtract $n^2$ from the whole inequality? $\endgroup$ – Loo Soo Yong May 22 at 15:40
  • $\begingroup$ yes, you can subtract the term $n^2$ from the whole inequality. $\endgroup$ – Dr. Sonnhard Graubner May 22 at 15:47
  • $\begingroup$ We get in this case $$0<2n<2n+1$$ which is true for $$n>0$$ $\endgroup$ – Dr. Sonnhard Graubner May 22 at 15:49
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$n(n+2)-n^2=2n>0$

$(n+1)^2-n(n+2)=1>0$

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Because $n> 0$

$$n^2 + 2n > n^2$$ trivially.

$$(n+1)^2 = n^2 + 2n + 1 > n^2 + 2n$$

Therefore,

$$n^2 < n(n+2) < (n+1)^2$$

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