$C=\{\frac{n^2+12n+32}{n+5}\:|\:n\in \mathbb{N}\}$. Prove that $\inf C=\frac{45}{6}$

Question

Define $C=\{\frac{n^2+12n+32}{n+5}\:|\:n\in \mathbb{N}\}$. Prove that $\inf C=\frac{45}{6}$

Attempt:
We'll show that $\frac{45}{6}$ is a lower bound of $A$. Let $c\in C$. There exists $n\in \mathbb{N}$ s.t. $c=\frac{n^2+12n+32}{n+5}$.
$$\frac{45}{6}\leq\frac{n^2+12n+32}{n+5}\iff...\iff11\leq2n^2+9n$$To prove that $\frac{45}{6}$ is infimum, it is sufficient to show $\forall \epsilon>0\quad \exists c\in C\quad(c<\epsilon+\inf C) $.
Using $$\frac{n_0^2+12n_0+32}{n_0+5}<\epsilon+\frac{45}{6}$$I wasn't able to find such $n_0\in \mathbb{N}$ that would do the job.

Answer 1

Let $f(x)=\frac{x^2+12x+32}{x+5}=x+7-\frac3{x+5}$. Then $f'(x)=1+\frac3{(x+5)^2}>0$. So, $f$ is strictly increasing and therefore$$(\forall n\in\mathbb{N}):f(n)\geqslant f(1)=\frac{45}6.$$

Answer 2

$$\iff n^2+n(12-C)+32-5C=0$$ which is a quadratic equation in $n$

As $n$ is real, $$(12-C)^2\ge4(32-5C)\iff21C^2-24C+16\ge0$$

Answer 3

To answer to the question raised in a comment, it can be solved just using properties of inequalities: in the decomposition of $$ f(n)=\frac{n^2+12n+32}{n+5}=n+7 -\frac 3{n+5}, $$ note $n+5$ is increasing, so as it is positive for all $n\in\mathbf N$, $\dfrac 3{n+5}$ is decreasing, and $-\dfrac 3{n+5}$ is (again) increasing.

Thus $f(n)$, which is the sum of two increasing functions, is increasing.

Edit: B.t.w., as $\mathbf N$ contains $0$, the minimum should be $\dfrac{32}5$.