# Prove that $n(n+2)$ lies between $n^2$ and $(n+1)^2$, given n is a positive integer

I'm currently trying to prove the inequality $$n^2

Is it possible to solve this without induction?

• Just expand them. – Theo Bendit May 22 at 15:35
• So I got $n^2<n^2 +2n<n^2 +2n+1$ and $0<2n<2n+1$ by subtracting $n^2$ is this correct? – Loo Soo Yong May 22 at 15:37
• 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. – Theo Bendit May 22 at 15:39

Expanding we get $$n^2 Can you proceed?

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

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

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

Because $$n> 0$$

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

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

Therefore,

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