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: 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 $\frac{243}{32}$.
A: Let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.
Hence, the condition does not depend on $w^3$ and we need to find a maximal value of $f,$ where
$$f(w^3)=27u^3-27uv^2+7w^3.$$
We see that $f$ increases, 
which says that it's enough to solve our problem for the maximal value of $w^3$.
Now, $p$, $q$ and $r$ are three non-negative roots of the following equation.
$$(x-p)(x-q)(x-r)=0$$ or
$$x^3-3ux^2+3v^2x-w^3=0$$ or
$$x^3-3ux^2+3v^2x=w^3.$$
Let $u$ and $v^2$ will be constants and $w^3$ is changing. 
Since the line $y=w^3$ and the graph of $y=x^3-3ux^2+3v^2x$ have three common points,
we see that $w^3$ will get a maximal value, 
when a line $y=w^3$ is a tangent line to the graph of $y=x^3-3ux^2+3v^2x$,
which happens for equality case of two variables.
Also, we need to check the case, when one of our variables is equal to $\frac{3}{2}.$


*

*Two variables are equal.


Since our inequality is symmetric, we can assume $b=a$.
Thus, $c=3-2a\leq\frac{3}{2},$ which gives $\frac{3}{4}\leq a\leq\frac{3}{2}$ and we need to find a maximum of $g$, where
$$g(a)=2a^3+(3-2a)^3+4a^2(3-2a)$$  and easy to see that
$$\max_{\frac{3}{4}\leq a\leq \frac{3}{2}}g=g\left(\frac{3}{4}\right)=\frac{243}{32}.$$
2. One of variables is equal to $\frac{3}{2}.$
Let $c=\frac{3}{2}.$
Hence, $b=\frac{3}{2}-a$ and we need to find a maximal value of $h$, where 
$$h(a)=a^3+\left(\frac{3}{2}-a\right)^3+\frac{27}{8}+3a(3-2a)$$ and easy to see that $$\max_{0\leq a\leq\frac{3}{2}}h=h\left(\frac{3}{4}\right)=\frac{243}{32},$$
which gives the answer: $\frac{243}{32}.$
