<s>[INST] 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.
 [/INST] 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: $\frac{243}{32}.$
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