Cyclic Algebra. Suppose that $a,b,c$ are real numbers satisfying $a^2 + b^2 + c^2 = 1$ and $a^3 + b^3 + c^3 = 1$.
Find all possible value(s) of $a+b+c$.
My solution :
$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow   a+b+c=1+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow a+b+c=1+c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)$
$\Rightarrow   a+b+c=1+c(1-c^2)+a(1-a^2)+b(1-b^2)$
$\Rightarrow   a+b+c=1+c-c^3+a-a^3+b-b^3$
$\Rightarrow   a+b+c=1+c+a+b-1$
$\Rightarrow   a+b+c=c+a+b$
This led me to the same expression $a+b+c$.
Is there a smarter way to solve this problem?

For a collection $a_1,...,a_n$ of real numbers, let $m=\max|a_i|$. We notice that $m \leq \big(\sum_i |a_i|^2\big)^{1/2}$ with equality iff there exists an index $i$ such that $|a_i| = m$ and $a_j=0$ for $j \neq i$. Moreover, since $|a_i|^3 \leq m|a_i|^2$ we notice that $$\bigg( \sum_i |a_i|^3 \bigg)^{1/3} \leq \bigg( \sum_i |a_i|^2 \bigg)^{1/3} \cdot m^{1/3} \leq \bigg( \sum_i |a_i|^2 \bigg)^{1/3} \cdot \bigg(\sum_i |a_i|^2 \bigg)^{ \frac12 \cdot \frac13} = \bigg( \sum_i |a_i|^2 \bigg)^{1/2}.$$ Now if we know that $\sum a_i^2 = \sum a_i^3 =1$, then it holds that $$1 = \bigg(\sum a_i^3 \bigg)^{1/3} \leq \bigg( \sum |a_i|^3 \bigg)^{1/3} \leq \bigg( \sum a_i^2 \bigg)^{1/2} = 1,$$ and therefore equality must hold throughout both of the previous expressions. This forces one to have $m= \big(\sum a_i^2 \big)^{1/2}$. Hence some $|a_i|=m$ and all other $a_j$ are zero. This forces $a_i=1$, hence the sum must be $1$.
