The minimum seems to be easier

This is just a constrained optimization problem with a Lagrangian of $\large \mathcal{L}(a,b,c)=a^{3}+b^{3}+c^{3}-4\cdot(a^{2}+b^{2}+c^{2}) - \lambda\cdot(a+b+c-3)$ .

We then have $\large \nabla\mathcal{L}=0$ :

$\large \frac{d\mathcal{L}}{da}=3a^{2}-8a-\lambda=0$

$\large \frac{d\mathcal{L}}{db}=3b^{2}-8b-\lambda=0$

$\large \frac{d\mathcal{L}}{dc}=3c^{2}-8c-\lambda=0$

$\large \frac{d\mathcal{L}}{d\lambda}=-(a+b+c-3)=0$

Solving for $a$ , $b$ and $c$ gives three possibilities:

$a=b=c=1\vee a=b=\frac{1}{3}, c=\frac{7}{3} \vee a=b=\frac{4}{3},c=\frac{1}{3}$ , which when plugged in give respectively $-9,-\frac{89}{9},-\frac{89}{9}$ .

The maximum is then $-9$ , for $a=b=c=1$ .

Side note: I'm not sure why the title says the minimum is easier to find.

Let's define a function $f(x)$ as :

$f(x)=x^3-4x^2$

$f'(x)=3x^2-8x \implies f''(x)=6x-8$

So, $f(x)$ is convex for $x<\frac{4}{3}$

By using Jensen's inequality in the interval $(0,\frac{3}{4})$ , we get :

$f(a)+f(b)+f(c) \le 3f(\frac{a+b+c}{3})$

$a^3+b^3+c^3-4(a^2+b^2+c^2) \le 3(1-4) =-9$

Equality holds when $a=b=c=1$ .

I found it helpful to draw a graph of $f(x)=x^3-4x^2$ . Let $S=f(a)+f(b)+f(c)$

$f(x) \leq 0$ for $x \in [0,3]$

WLOG $a \leq b \leq c$

If $c>1$ , then $f(c) \leq -9$ , so $S \leq -9$ .

If $c \leq 1$ then $a=b=c=1$ , So $S=-9$

Therefore $-9$ is the maximum.

We know that a, b and c are positive reals and their sum is equal to 9. Therefore, the only solutions to that is: (0,0,3), (1,1,1) and (2,1,0). After substituting those valuers, the maximum is -9.

Note: sorry for my bad grammar.