We can expand the left-hand side of the inequality as follows: $(a-b)(a^2-9)+(a-c)(b^2-9)+(b-c)(c^2-9) = a^2(a-b)-9(a-b)+a(b^2-9)-c(b^2-9)+b(c^2-9)-c^2(b-c)$ This expression can be simplified as: $a^2(a-b)+b^2(b-c)+c^2(c-a)$ Now, we can apply the Trivial Inequality to each term: $a^2(a-b) + b^2(b-c) + c^2(c-a) \le (a-b)^2 + (b-c)^2 + (c-a)^2$ This reduces to: $(a-b)^2 + (b-c)^2 + (c-a)^2 \le 36$ Since all the variables are non-negative, we have: $(a-b) + (b-c) + (c-a) \le 6$ This means that $|a-b| + |b-c| + |c-a| \le 6$. Since the absolute value of each term is non-negative, we can drop the absolute values and get: $a-b + b-c + c-a \le 6$ This simplifies to: $a + c \le b + 6$ Since $a \le b$ and $b \le c$, we have: $a + c \le 2b \le b + 6$ Thus, $b \ge 3$, which means that $b = 3$. Substituting this back into the inequality, we get: $(a-3)(a^2-9)+(a-c)(9-9)+(3-c)(c^2-9) \le 36$ This reduces to: $(a-3)(a^2-9) \le 36$ Since $a \le 3$, the inequality is satisfied for all possible values of $a$. Therefore, we have proved that $(a-b)(a^2-9)+(a-c)(b^2-9)+(b-c)(c^2-9) \le 36$ for all non-negative values of $a$, $b$, and $c$ such that $0 \le a \le b \le c \le 3$.

We can expand the left-hand side of the inequality as follows: $(a-b)(a^2-9)+(a-c)(b^2-9)+(b-c)(c^2-9) = a^2(a-b)-9(a-b)+a(b^2-9)-c(b^2-9)+b(c^2-9)-c^2(b-c)$ This expression can be simplified as: $a^2(a-b)+b^2(b-c)+c^2(c-a)$ Now, we can apply the Trivial Inequality to each term: $a^2(a-b) + b^2(b-c) + c^2(c-a) \le (a-b)^2 + (b-c)^2 + (c-a)^2$ This reduces to: $(a-b)^2 + (b-c)^2 + (c-a)^2 \le 36$ Since all the variables are non-negative, we have: $(a-b) + (b-c) + (c-a) \le 6$ This means that $|a-b| + |b-c| + |c-a| \le 6$. Since the absolute value of each term is non-negative, we can drop the absolute values and get: $a-b + b-c + c-a \le 6$ This simplifies to: $a + c \le b + 6$ Since $a \le b$ and $b \le c$, we have: $a + c \le 2b \le b + 6$ Thus, $b \ge 3$, which means that $b = 3$. Substituting this back into the inequality, we get: $(a-3)(a^2-9)+(a-c)(9-9)+(3-c)(c^2-9) \le 36$ This reduces to: $(a-3)(a^2-9) \le 36$ Since $a \le 3$, the inequality is satisfied for all possible values of $a$. Therefore, we have proved that $(a-b)(a^2-9)+(a-c)(b^2-9)+(b-c)(c^2-9) \le 36$ for all non-negative values of $a$, $b$, and $c$ such that $0 \le a \le b \le c \le 3$.