4. Given numbers $a, b, c \in [0, \infty)$, such that $a+b+c=3$. Determine the extreme values of the expression $2\left(a^{2}+b^{2}+c^{2}\right)+3 a b c$.

Note. All subjects are mandatory and are graded with scores ranging from 0 to 7 points. Time for work, three hours.

## Grading Rubric
4. Given numbers $a, b, c \in [0, \infty)$, such that $a+b+c=3$. Determine the extreme values of the expression
$2\left(a^{2}+b^{2}+c^{2}\right)+3 a b c$

## Grading Rubric.

We prove that $9 \leq 2\left(a^{2}+b^{2}+c^{2}\right)+3 a b c \leq 18$, with equality on the left for $a=b=c=1$, and on the right for $a=3, b=c=0$, or $b=3, a=c=0$, or $c=3, a=b=0$.

The right inequality can be written as $2(a+b+c)^{2}-4(a b+b c+c a)+3 a b c \leq 18 \Leftrightarrow 3 a b c \leq 4(a b+b c+c a)$, which is true because $a, b, c \in [0,3]$ and $3 a b c = a b c + a b c + a b c \leq 3 a b + 3 b c + 3 a c \leq 4(a b + b c + c a)$ (3 points). Equality holds for $a b + b c + c a = 0$, i.e., $a=b=0$, or $b=c=0$, or $c=a=0$, the case $a=b=c=0$ being excluded by the condition in the problem (1 point).

The left inequality can be written as $2(a+b+c)^{2}-4(a b+b c+c a)+3 a b c \geq 9 \Leftrightarrow 4(a b+b c+c a)-3 a b c \leq 9$. To prove this, assume $a \leq b \leq c$; in this case, we have $a \leq 1$ and the inequality to be proved becomes $4 a(b+c) + 4 b c - 3 a b c \leq 9 \Leftrightarrow 4 a(3-a) + b c(4-3 a) \leq 9$. Since $4-3 a \geq 0$, we have $b c(4-3 a) \leq \frac{(b+c)^{2}(4-3 a)}{4} = \frac{(3-a)^{2}(4-3 a)}{4}$, and the inequality to be proved becomes $4 a(3-a) + \frac{(3-a)^{2}(4-3 a)}{4} \leq 9$, which is equivalent, after some calculations, to $3 a(a-1)^{2} \geq 0$ (2 points). Equality holds for $a=b=c=1$ (1 point).