Half-trivial isn't that trivial!

Let $S = a + ab + abc$

As the other solution has mentioned, it is apparent that the minimum value of $S$ is $0$ since $a, b, c \geq 0$

I have another approach to find the maximum of $S$

$S = a(1 + b + bc)$

Substituting $a = 3 - b - c$

$S = (3 - b - c)(1 + b + bc)$

$\Leftrightarrow S = 3 - b - c + 3b - b^{2} - bc + 3bc - b^{2}c - bc^{2}$

$\Leftrightarrow S = -c^{2}b - c(b^{2} - 2b + 1) - (b^{2} - 2b + 1) + 4$

$\Leftrightarrow S = -[c^{2}b + (c+1)(b-1)^{2}] + 4$

Since $c^{2}b \geq 0$ and $(c+1)(b-1)^{2} \geq 0$

$\Rightarrow S \leq 4$

The inequality happens when $c^{2}b = (c+1)(b-1)^{2} = 0$ , or $a = 2, b = 1, c = 0$

$\Rightarrow$ Sum of the maximum and minimum value of $S = a + ab + abc$ is $4$

Relevant wiki: Lagrange Multipliers

Since $a,b,c \geq 0$ , $a+ab+abc$ can't be negative and its minimum value is $0$ (that occurs when $a=0$ )

Now we are going to search the maximum value using the Lagrange multipliers method:

$\mathcal{L}(a,b,c,\lambda)=a+ab+abc-\lambda(a+b+c-3)$

Taking the partial derivatives with respect to $a$ , $b$ and $c$ and setting them equal to $0$ we get

$\begin{cases} 1+b+bc=\lambda \\ a+ac=\lambda \\ ab=\lambda \end{cases}$

Substituting $\lambda=ab$ in the second equation and simplifying $a$ (that can't be $0$ since that would give us the minimum)

$a+ac=ab \iff b=c+1$

Substituting $b=c+1$ in the first equation

$1+c+1+(c+1)c=\lambda \iff (c+1)^2+1=\lambda$

Since $\lambda=a(c+1)$ from the second equation, substituting it in the first we get a quadratic in $c+1$

$(c+1)^2-a(c+1)+1=0$

Solving for $c+1$

$c+1=\dfrac{a\pm\sqrt{a^2-4}}{2} \iff 2c=a\pm\sqrt{a^2-4}-2$

Substituting $b=c+1$ and $2c=a\pm\sqrt{a^2-4}-2$ into the original constraint $a+b+c=3$

$\begin{aligned} & a+c+1+c=3\\ \iff \quad & a+2c=2\\ \iff \quad & 2a\pm\sqrt{a^2-4}=4 \end{aligned}$

Solving separately the $+$ and $-$ cases we get from both $a=2$ (the $-$ case also gives $a=\dfrac{10}{3}$ , but we have to discard it because it would make $b$ or $c$ negative), which implies $c=\dfrac{2\pm\sqrt{2^2-4}-2}{2}=0$ and $b=0+1=1$ , therefore the maximum value of $a+ab+abc$ is $2+2\cdot 1+2\cdot 1\cdot 0=4$

In conclusion the sum of the maximum and minimum values of $a+ab+abc$ is

$4+0=\boxed{4}$