Prove by induction $n!<4n^2+3$ How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?
Here´s my try:
The equation is valid for every $n\leq4$.
n=5: n!=5!=120>103
n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$
Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.
Any ideas how i could continue or whats wrong?
 A: Your work is mostly fine, but you can perfect your redacting.
What you want to show is that the set $A=\{n\in\mathbb N\mid n!<4n^2+3\}=\{0,1,2,3,4\}$
For that you need to verify the inequality for the a few numbers which you certainly did, but it needs to appear on your solution.
$\begin{cases}
n=0&0!=1<4\times 0^2+3=3&\text{ok}\\
n=1&1!=1<4\times 1^2+3=7&\text{ok}\\
n=2&2!=2<4\times 2^2+3=19&\text{ok}\\
n=3&3!=6<4\times 3^2+3=39&\text{ok}\\
n=4&4!=24<4\times 4^2+3=67&\text{ok}\\
n=5&5!=120\not<4\times 5^2+3=103&\text{not verified}\\
\end{cases}$
You are correctly suspecting that the inequality is not verified either for $n\ge 5$, so you should explicit the induction proposition :
$$P(n) : n! \ge 4n^2+3\qquad\forall n\ge 5$$
Base case $P(5)$ is verified.
So let assume $P(n)$ we get $(n+1)!=(n+1)n!\ge (n+1)(4n^2+3)$
And this is where you got stuck.
You need to make appear $4(n+1)^2+3$, instead I propose to subtract both quantities and show the difference is positive (this is easier to handle):
$\begin{align}(n+1)!-\bigg((4(n+1)^2+3\bigg)
&\ge (n+1)(4n^2+3)-\bigg(4(n+1)^2+3\bigg)
\\&\ge 4n^3+3n+4n^2+3-(4n^2+8n+7)
\\&\ge 4n^3-5n-4
\\&\ge 4n^3-5n-4n\qquad\text{ for } n\ge 1\implies 4\le 4n
\\&\ge n(4n^2-9)
\\&\ge 0 \qquad\qquad\qquad\qquad\text{ for } n^2\ge \frac 94\iff n\ge \frac 32
\end{align}$
Since $n\ge 5$ both conditions ($n\ge 1$ and $n\ge \frac 32$) are verified therefore you have proved your induction step.
A: Here is a variant without induction.
For $n\ge 5\ $ then $\, n!=n(n-1)\underbrace{(n-2)(n-3)}_{\ge 6}\underbrace{\cdots 1}_{\ge 1}\ge 6n(n-1)$
Let $f(n)=6n(n-1)-(4n^2+3)=2n^2-6n-3$
$f'(n)=4n-6$ so $f\ \nearrow$ for $n\ge \frac 32$ and since $f(5)=17>0$ then $n!>4n^2+3$ for $n\ge 5$
A: Assuming you want to prove that for each $n\geq 5$ one has $n! > 4n^2+3$ just note that it easily holds for $n=5$. Now assume that it holds up to $n \geq n-1 \geq \ldots 5.$ Then
$$(n+1)! = (n+1)n! > (n+1)(4n^2+3).$$
It is only left to prove that $(n+1)(4n^2+3) > 4(n+1)^2+3$ for each $n\geq 5.$ In fact,
$$(n+1)(4n^2+3)-4(n+1)^2-3 = 4n^3-5n-4.$$
So let $p(n) = 4n^3-5n-4.$ Note that $p(5) > 0.$ Moreover, $p'(n) = 12n-5 > 0,~\forall n\geq 5.$
