<s>[INST] Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$ Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$
Case $S(1)$ is true:
$$2((1)+2)\leq((1)+2)^{2}$$
$$6\leq9$$
Case $S(n)$ is true for all $n=1,2,...$
$$2(n+2)\leq(n+2)^{2}(i)$$
Case $S\left(n+1\right)$
$$2(n+3)\leq(n+3)^{2}(ii)$$
From (i)
$$2(n+2+1-1)\leq(n+2+1-1)^{2}$$
$$2(n+3)-2(1)\leq(n+3)^{2}-2(n+3)+1$$
$$2(n+3)\leq(n+3)^{2}-2(n+3)+3$$
$$2(n+3)\leq(n+3)^{2}-(2n+3)$$
$$2(2n+3)+3\leq(n+3)^{2}$$
Thus $(i)$ is true for all $n=1,2,...$
My question: How can I get the same expression as (ii)?
I got to $2(2n+3)+3\leq(n+3)^{2}$ but it's clearly wrong
 [/INST] Note that$$2(n+2)-2(n+1)=2$$and that$$(n+3)^2-(n+2)^2=2n+5.$$So, if $2(n+1)\leqslant(n+2)^2$, you have\begin{align}2(n+2)&=2(n+1)+2\\&\leqslant(n+2)^2+2\\&\leqslant(n+2)^2+2n+5\\&=(n+3)^2.\end{align}
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