Maximizing $\frac{a^2+6b+1}{a^2+a}$, where $a=p+q+r=pqr$ and $ab=pq+qr+rp$ for positive reals $p$, $q$, $r$ 
Given $a$, $b$, $p$, $q$, $r \in\mathbb{R_{>0}}$ s.t.
  $$\begin {cases}\phantom{b}a=p+q+r=pqr \\ab =pq+qr+rp\end{cases} $$
  Find the maximum of $$\dfrac{a^2+6b+1}{a^2+a}$$


This question is terrifying that I even don't know how to start it. I've found different ways such as multiplying the numerator and denominator by $a$ and substitute, but it is not useful (maybe?). Anyway, please comment or answer if you solve it or having clues that may help solve this question.

Recently, I find a solution which is quite simple. Check it out!

Claim $1$: $a^2\ge3ab\rightarrow a\ge3b, b\le\dfrac{a}{3}$

Prove:
$$a^2-3ab=\left(x+y+z\right)^2-3\left(xy+yz+zx\right)=x^2+y^2+z^2-xy-yz-zx=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{2}\ge0 \\ \therefore a^2\ge 3ab$$

Claim $2$: $a\ge3\sqrt{3}$

Prove:
By AM-GM Inequality,
$$\left(\dfrac{x+y+z}{3}\right)^3\ge xyz\\\left(\dfrac{a}{3}\right)^3\ge a\\\dfrac{a^3}{27}\ge a\\a^2\ge27\\\therefore a\ge3\sqrt{3}$$
Then comes the solution:
$\dfrac{a^2+6b+1}{a^2+a}\le\dfrac{a^2+2a+1}{a^2+a}=\dfrac{\left(a+1\right)^2}{a\left(a+1\right)}=\dfrac{a+1}{a}=1+\dfrac{1}{a}\le1+\dfrac{1}{3\sqrt{3}}=\boxed{1+\dfrac{\sqrt{3}}{9}}$
Equality holds when $x=y=z=b=\sqrt{3},a=3\sqrt{3}$
