The minimum possible value is $\dfrac{\sqrt{3}}{3}$, while the maximum one is $2$. For the minimum value, note that $a^3+b^3+c^3 \geq \dfrac{(a+b+c)^3}{9} \geq \dfrac{(\sqrt{3(ab+bc+ca)})^3}{9}=\dfrac{\sqrt{3}}{3},$ with equality when $a=b=c=\dfrac{\sqrt{3}}{3}$. For the maximum value, note that $a^3+b^3+c^3 \leq a+b+c,$ and so we only have to prove that $a+b+c \leq 2$. Since $0 \leq a,b,c \leq 1$, we obtain that $(a-1)(b-1) \geq 0,$ that is $ab \geq a+b-1$, and adding similar expressions we obtain $1=ab+bc+ca \geq 2(a+b+c)-3,$ hence $a+b+c \leq 2$, as desired. Equality holds, for example, when $(a,b,c)=(1,1,0)$.

Let $a, b, c \in [0;1]$ and $ab+bc+ca=1$. Prove that$$\sqrt{3^{2-k}} \leq a^k+b^k+c^k\leq 2$$Where $3\leq k\in N^+.$ Let $a, b, c \in [0;1]$ and $a^2+b^2+c^2=1$. Prove that$$\frac{\sqrt{3}}{3}\leq a^3+b^3+c^3\leq 1$$

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Let $a, b, c \geq 0$ and $a^3+b^3+c^3=1$. Prove that$$ a^2+b^2+c^2 \leq \sqrt[3]{3}$$

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