Given the positive real numbers $0\le a,b,c\le 2$ and $a+b+c=3$. Prove that $a^3+b^3+c^3\le 9$

Question

Given the positive real numbers $$0\le a,b,c\le 2$$ and $$a+b+c=3$$. Prove that $$a^3+b^3+c^3\le 9$$

Answer 1

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 $a=2, b=1, c=0$.

Answer 2

Let $a\geq b\geq c$.

Since $f(x)=x^3$ is a convex function on $[0,3]$ and $(2,1,0)\succ(a,b,c)$,

by Karamata we obtain $$9=2^3+1^3+0^2\geq a^3+b^3+c^3$$ and we are done!