Prove that the following inequality holds for all $n \in N$ by using the principle of mathematical induction:
$(2n + 7) < (n + 3)^{2}$

(N/A) Let the given statement be $P(n)$,i.e.,
$P(n): (2n + 7) < (n + 3)^{2}$
Step $1$: Check for $n = 1$.
$2(1) + 7 = 9$ and $(1 + 3)^{2} = 16$.
Since $9 < 16$,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some positive integer $k$,i.e.,
$(2k + 7) < (k + 3)^{2}$ ..........$(i)$
Step $3$: We shall now prove that $P(k + 1)$ is true whenever $P(k)$ is true.
Consider the expression for $n = k + 1$:
$2(k + 1) + 7 = (2k + 7) + 2$
Using $(i)$,we have:
$2(k + 1) + 7 < (k + 3)^{2} + 2$
$2(k + 1) + 7 < k^{2} + 6k + 9 + 2$
$2(k + 1) + 7 < k^{2} + 6k + 11$
Since $k^{2} + 6k + 11 < k^{2} + 8k + 16$ for all $k \in N$ (because $2k + 5 > 0$ for $k \geq 1$),
$2(k + 1) + 7 < (k + 4)^{2}$
$2(k + 1) + 7 < \{(k + 1) + 3\}^{2}$
Thus,$P(k + 1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N$.