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## Using the principle of mathematical induction to show inequalities I

Solution:

### Part a

The $n$-th proposition is
$$P_n: \quad n^2>n+1.$$ Clearly, $P_2$ is true because $2^2=4$ which is greater than $2+1=3$. We have the induction basis.

Now we assume $P_n$ is true (here we assume that $n\ge 2$), that is $n^2>n+1$. We would like to show $P_{n+1}$ is true based on $P_n$. Note that $n>0$, we have
\begin{align*}
(n+1)^2=n^2+2n+1>n+1+2n+1>(n+1)+1.
\end{align*} Therefore $P_{n+1}$ is true if $P_n$ is true. By the principle of mathematical induction, we make the conclusion that $P_n$ is true for all positive integers $n$.

### Part b

The $n$-th proposition is
$$P_n: \quad n!> n^2.$$ Clearly, $P_4$ is true because $4!=24$ which is greater than $4^2=16$. We have the induction basis.

Now we assume $P_n$ is true (here we assume that $n\ge 4$), that is $n!> n^2$. We would like to show $P_{n+1}$ is true based on $P_n$. Note that $n\ge 4$, we have
\begin{equation}\label{eq:1-8-1}
n^2\ge 4n>n,\quad n^2 >1.
\end{equation} Therefore
\begin{align*}
(n+1)!=&\ (n+1)\cdot n!>4\cdot n^2\\
=&\ n^2+2n^2+n^2\\
\text{use \eqref{eq:1-8-1}}\quad >&\ n^2+2n+1=(n+1)^2.
\end{align*} Therefore $P_{n+1}$ is true if $P_n$ is true. By the principle of mathematical induction, we make the conclusion that $P_n$ is true for all positive integers $n$.