<s>[INST] A constrained extremum problem Find the maximum possible value of $$A = a^{333} + b^{333}+c^{333}$$ subject to the constraints $$a+b+c=0$$ and $$a^2+b^2+c^2=1,$$
where $a,b,c\in \mathbb{R}$
Thank you for helping me.
 [/INST] let $A_{n}=a^n+b^n+c^n$ then we have
$$A_{n+3}=(a+b+c)A_{n+2}-(ab+bc+ac)A_{n+1}+abcA_{n}$$
and 
$$ab+bc+ac=-\dfrac{1}{2}$$
so
$$A_{n+3}=\dfrac{1}{2}A_{n+1}+abcA_{n}$$
and use $$a+b+c=0,\Longrightarrow A_{3}=a^3+b^3+c^3=3abc$$
and $$A_{1}=0,A_{2}=1,A_{3}=3abc$$
so
$$A_{4}=\dfrac{1}{2}A_{2}+abcA_{1}=\dfrac{1}{2}$$
$$A_{5}=\dfrac{1}{2}A_{3}+abcA_{2}=\dfrac{3}{2}abc+abc=\dfrac{5}{2}abc$$
$$A_{6}=\dfrac{1}{2}A_{4}+abcA_{3}=\dfrac{1}{4}+3(abc)^2$$
$$A_{7}=\dfrac{1}{2}A_{5}+abcA_{4}=\dfrac{5}{4}abc+\dfrac{1}{2}abc=\dfrac{7}{4}abc$$
$$A_{8}=\cdots, A_{9}=\cdots,\cdots$$
$$A_{300}=f(abc)$$
and we have $|abc|\le\dfrac{1}{\sqrt{54}}$,see:$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$
</s>