Maximizing a cyclic polynomial.

There are two cases: $a, b, c$ are all positive reals, or one of them is equal to zero.

Suppose $a, b, c$ are all non-zero. Assume WLOG that $a \geq b \geq c$ . Then using rearrangement inequality on the sets $\{ab, ac, bc\}$ and $\{a, b, c\}$ , $ab^2 + bc^2 + ca^2 \geq 3abc$ . Using AM-GM, we have that $1 = a+b+c \geq 3 \sqrt[3]{abc}$ , so $1 \geq 27abc$ and $3abc \leq \frac{1}{9}$ . For the rearrangement inequality, equality is reached when $a=b=c$ , and for AM-GM equality is reached when $a=b=c$ , so this result is both obtainable and maximized.

Suppose $c$ is equal to zero. Then $a+b = 1$ and we want to maximize $ab^2$ . By AM-GM, $a+\frac{b}{2}+\frac{b}{2} \geq 3\sqrt[3]{\frac{ab^2}{4}}$ . Cubing both sides, we get $(a+b)^3 = 1 \geq \frac{27}{4} ab^2$ , so $ab^2$ is maximized at $\frac{4}{27}$ . So $\boxed{\frac{4}{27}}$ is our answer.

Critical point is where the gradient is perpendicular to the plane $a+b+c=1$ . This happens where $a=b=c$ . Value is $1/9$ . When checking the border off the plane, all three line sections have a maximum of $4/27$ .

$$ I have to admit it, that this is the tricky one question. My first approach is using Lagrange multiplier since I hate using AM-GM inequality although It's sometimes useful for certain case. My first attempt was $10$ because the maximum value that I found by using Lagrange multiplier was $\frac{1}{9}$ . But Brilliant said "my answer is incorrect". I checked my approach again, but it seemed nothing was wrong. So, I read the question carefully until I found this term: "non-negative real numbers". Truth be told, I'm not that good in math theory. Mostly, all of the problem that I've solved here by intuitive method. The definition of non-negative number is a number that is either 0 or positive. So I guessed that at least one of $\,a$ , $\,b$ , or $\,c$ maybe be 0. In this case, it's not really important which one of them is equal to 0 because the form of the equation will be 'similar'. Let's say $\,c=0$ then the question turns out to be $$ $\text{max} \{\left.ab^2 \right| a+b=1\}.$ $$ Once again I used Lagrange multiplier and found $$ $\text{max} \{\left.ab^2 \right| a+b=1\}=\frac{4}{27}, \;\;\; \text{where}\;a=\frac{1}{3}\; \text{and}\;b=\frac{2}{3}.$ $$ Thus, $\,m+n=\boxed{31}$ . Luckily, in the second attempt my answer was correct . If not, I wouldn't take any risk to try the third attempt. Hahaha... $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Case I : WLOG, $a,b=0, c=1$

Case II: WLOG, $a=0, b,c>0$ Maximum $\displaystyle ab^2+bc^2+ca^2=\frac{4}{27}$

Case III: $a,b,c$ Maximum $\displaystyle ab^2+bc^2+ca^2=\frac{1}{9}$

and $\displaystyle \frac{4}{27}>\frac{1}{9}$

$n+m=31$