Finding the maximum of $p^3 + q^3 +r^3 + 4pqr$

Question

$p$,$q$,$r$ 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 $p^3 + q^3 + r^3 + 4pqr$ ?

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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.

Answer 1

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 $p$.

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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.

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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 $\frac{243}{32}$.

Answer 2

To maximize $$ p^3+q^3+r^3+4pqr\tag1 $$ while maintaining $$ p+q+r=3\tag2 $$ Means that we need to find $p,q,r$ so that for all variations so that $$ \delta p+\delta q+\delta r=0\tag3 $$ we also have $$ \left(3p^2+4qr\right)\delta p+\left(3q^2+4pr\right)\delta q+\left(3r^2+4pq\right)\delta r=0\tag4 $$ Orthogonality requires that $$ 3p^2+4qr=3q^2+4pr=3r^2+4pq\tag5 $$ If $p\ne q$, then $$ \begin{align} 3\left(p^2-q^2\right)&=4r(p-q)\\ \frac34(p+q)&=r\tag6 \end{align} $$ Since $p\ne q$, either $p\ne r$ or $q\ne r$. WLOG assume $p\ne r$. Then, as in $(6)$, we have $$ \frac34(p+r)=q\tag7 $$ Comparing $(6)$ and $(7)$ yields $$ 3p=q=r\tag8 $$

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Noting symmetries, and the solution $p=q=r$, we get the solutions $$ \left\{(1,1,1),\left(\frac37,\frac97,\frac97\right),\left(\frac97,\frac37,\frac97\right),\left(\frac97,\frac97,\frac37\right)\right\}\tag9 $$ which give the values in $(1)$ to be $$ \left\{7,\frac{351}{49},\frac{351}{49},\frac{351}{49}\right\}\tag{10} $$ The point $(1,1,1)$ is a local minimum and the other three are saddle points.

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There are also the edge cases of $p=0$ or $q=0$ or $r=0$. WLOG assume $r=0$. Equations $(5)$ become $p=q$. Symmetry gives the cases $$ \left\{\left(\frac32,\frac32,0\right),\left(\frac32,0,\frac32\right),\left(0,\frac32,\frac32\right)\right\}\tag{11} $$ which give the values in $(1)$ to be $$ \left\{\frac{27}4,\frac{27}4,\frac{27}4\right\}\tag{12} $$

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Then there are the corner cases $$ \{(3,0,0),(0,3,0),(0,0,3)\}\tag{13} $$ which give the values in $(1)$ to be $$ \{27,27,27\}\tag{14} $$

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Given the constraint that $p,q,r\le\frac32$, we have three more edge cases symmetric to $r=\frac32$ and $p+q=\frac32$. Equations $(5)$ become $3p^2+6q=3q^2+6p$, which, if $p\ne q$, implies $p+q=2$; thus, $p=q=\frac34$. Symmetry gives the cases $$ \left\{\left(\frac32,\frac34,\frac34\right),\left(\frac34,\frac32,\frac34\right),\left(\frac34,\frac34,\frac32\right)\right\}\tag{15} $$ which give the values in $(1)$ to be $$ \left\{\frac{243}{32},\frac{243}{32},\frac{243}{32}\right\}\tag{16} $$

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The greatest value inside the $p,q,r\le\frac32$ constraint is attained by case $(15)$. This gives the maximum of $$ \bbox[5px,border:2px solid #C0A000]{\frac{243}{32}}\tag{17} $$

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Here is a plot of $p^3+q^3+r^3+4pqr$ where $r=3-p-q$. The edges are where $p=\frac32$, $q=\frac32$, and $p+q=\frac32$. The points in $(15)$ are colored red. The interior points are those in $(9)$ and the corners are those in $(11)$.

Answer 3

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 $\frac{3}{2}.$

1. 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: $\frac{243}{32}.$