<s>[INST] Given the positive real numbers $0\le a,b,c\le 2$ and $a+b+c=3$. Prove that $a^3+b^3+c^3\le 9$ Given the positive real numbers $$0\le a,b,c\le 2$$ and $$a+b+c=3$$. Prove that $$a^3+b^3+c^3\le 9$$
 [/INST] Assume without loss of generality that $a$ is the maximum of $a,b,c$. This gives us that  $$ 3a \ge a+b+c=3 \iff 2 \ge a \ge 1 \tag{1}$$
From the condition $0 \le b,c \le a \le 2$. Note that we have that $$a^3+b^3+c^3 \le a^3+(b+c)^3=a^3+(3-a)^3=9\left(a-\frac{3}{2} \right)^2+\frac{27}{4} \tag{2}$$
Which follows from the fact that $b,c \ge 0$ and $ a+b+c=3$. 
Note that as we have $1 \le a \le 2$ from $(1)$ we have $$\left(a-\frac{3}{2}\right)^2 \le \frac{1}{4} \tag{3}$$
Thus  $$a^3+b^3+c^3 \le 9\left(a-\frac{3}{2} \right)^2+\frac{27}{4} \le \frac{9}{4}+\frac{27}{4}= 9 $$
From $(2)$ and $(3)$. Thus, our proof is done. Equality is true when $a=2, b=1, c=0$. 
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