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.

$a+b+c=0$, so at least two of $a,b,c$ have same sign, WLOG, let $bc \ge 0 $
$A=a^{333}+b^{333}+c^{333} = (-b-c)^{333}+b^{333}+c^{333}=- \sum_{i=1} ^{332} C_i b^{333-i}c^i$,
We want $A$ is max, so b,c must be negative, ,let $x=-b,y=-c $, $ \to x \ge 0,y \ge 0, xy \le \dfrac{1}{6}, x+y=\sqrt{\dfrac{1}{2}+xy} $
$ A=(x+y)^{333}-(x^{333}+y^{333}) \le \left(\dfrac{1}{2}+xy \right)^{\frac{333}{2}}-2(xy)^{\frac{333}{2}} $
$f(t)=\left(\dfrac{1}{2}+t \right)^k-2t^k$, we will prove $f(t)$ is mono increasing function when $ 0 \le  t \le \dfrac{1}{6} , k >3$:
$f'(t)=k\left[ \left(\dfrac{1}{2}+t \right)^{k-1}-2t^{k-1}  \right]$, $f'(t)>0 \iff \left(\dfrac{1}{2}+t \right)^{k-1}-2t^{k-1} >0 \iff \left(\dfrac{1}{2}+t \right)^{k-1}>2t^{k-1} \iff \left(\dfrac{1}{2}+t \right) >  2^{\frac{1}{k-1}}t \iff \dfrac{1}{2}> (p-1)t , p=2^{\frac{1}{k-1}} , 1<p<2, $ 
so last one is true, $f'(t)>0 \implies  A_{max}=\left(\dfrac{1}{2}+\dfrac{1}{6} \right)^{\frac{333}{2}}-2(\dfrac{1}{6})^{\frac{333}{2}}=\left(\dfrac{4}{6} \right)^{\frac{333}{2}}-2(\dfrac{1}{6})^{\frac{333}{2}}=\dfrac{2^{333}-2}{6^{166}\sqrt{6}}$, max will be hold when $x=y,xy=\dfrac{1}{6} \to x=y=\dfrac{1}{\sqrt{6}} \to b=c=-\dfrac{1}{\sqrt{6}}, a= \dfrac{2}{\sqrt{6}}$
