Finding the maximum of $p^3 + q^3 +r^3 + 4pqr$ 
$p$,$q$,$r$ are $3$ non-negative real numbers less than or equal to $1.5$ such that $p+q+r = 3$, what will be the maximum of $p^3 + q^3 + r^3 + 4pqr$ ?


I tried AM-GM on $p,q,r$ to get the maximum of $pqr$ as $1$, but on doing it for $p^3 + q^3 + r^3 $, I get the minimum, so I won't be able to combine them.
I tried to assume symmetry $p=q=r=1$, and the maximum value to be 7, but putting $p=1.1$, $q=1.1$ and $r=0.8$, I get the value to be $7.046 (> 7) $
How do I solve this with concepts like AM-GM?
PS: I do not know multivariable calculus concepts, however if it's not possible to solve it without using it, please show how to solve it with these concepts.

A hint before the answer:
$p^3+q^3+r^3+4pqr=p^3+(q+r)^3+4pqr-3(q+r)(qr)$
Try to then eliminate both $q$ and $r$ to make this in terms of $p$.

Motivation
Firstly one might try $p=q=r=1$ and get $p^3+q^3+r^3+4pqr=7$
But the trying of more cases will reveal that this is not the maximum. One might then try to fix a term and see what is the maximum with the fixed term. 
e.g. Set $p=3/2$. Then $p^3+q^3+r^3+4pqr=27/8+q^3+r^3+6qr$, $q+r=3/2$
The $q^3+r^3$ motivates writing $q^3+r^3+6qr$ in the form $(q+r)^3-something$ as $(q+r)^3=27/8$, dealing with the $q^3$ and $r^3$ terms.
i.e. $q^3+r^3+6qr=q^3+r^3+3(q+r)(qr)+(6-3q-3r)qr=(3/2)^3+(3/2)qr$
Then we are left to maximise $qr$, which is simple by AM-GM. (i.e. $q=r=3/4$) Trying to do this without fixing p at the start leads to the solution.

Solution
$$p^3+q^3+r^3+4pqr$$
$$=p^3+q^3+r^3+3(q+r)qr+(4p-3q-3r)qr$$
$$=p^3+(3-p)^3+(4p-(9-3p))qr$$
$$=p^3+(3-p)^3+(7p-9)qr$$
This is
less than or equal to $p^3+(3-p)^3+(7p-9)(\frac{3-p}{2})^2$, equality when $q=r$ if $p\geq 9/7$ and
less than or equal to $p^3+(3-p)^3+(7p-9)(1.5)(1.5-p)$, equality when one of $q, r$ equals $1.5$ if $p\leq 9/7$.
Note that among $p, q, r$, the smallest number has to be less than or equal to $1$. Without loss of generality $p\leq q \leq r$. Then $p\leq 9/7$. For $p^3+q^3+r^3+4pqr$ to be maximal, one of $q, r$ is 1.5. As $r$ is the largest, $r=1.5$.  
Finally, note that $3/2>9/7$, so we can apply the other case, replacing $r$ with $p$ (which is fine since $p^3+q^3+r^3+4pqr$ is symmetric over the 3 terms), concluding that if $r=3/2$, $p^3+q^3+r^3+4pqr$ is maximal when $p=q=3/4$.
Therefore, the maximum achievable is when one of $p, q, r$ is $3/2$ and the other 2 are $3/4$, to get a value of $\frac{243}{32}$.
