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: Somewhat against my initial expectations it seems to me that Lagrange multipliers work well this time.
The solution works for all exponents $n\ge3$, $n$ odd. Here $n=333$. At a critical point we have the vector equation
$$
\nabla(a^n+b^n+c^n)=\lambda_1\nabla (a^2+b^2+c^2-1)+\lambda_2\nabla(a+b+c).
$$
We see that $a,b,c$ are zeros of the polynomial
$$
P(x)=nx^{n-1}-2\lambda_1x-\lambda_2,
$$
because the above vector equation of gradients simplifies to
$$
(P(a),P(b),P(c))=(0,0,0).
$$
Here the graph $y=nx^{n-1}$ is convex (the second derivative is non-negative everywhere, and vanishes at an isolated point only). Therefore it can intersect the line $y=2\lambda_1x+\lambda_2$ at at most two distinct points. Irrespective of the values of $\lambda_1$ and $\lambda_2$.
As $P(x)=0$ has only two real solutions, we can conclude that at a critical point $a,b,c$ cannot all be distinct.
This allows us to find the critical points. For example, if $b=c$
we get from the constraints first the equations
$$
a+b+c=a+2b=0\qquad\text{and}\qquad a^2+b^2+c^2=a^2+2b^2=1.
$$
Then by solving $a=-2b$ from the first and plugging that into the second
that
$$
1=a^2+2b^2=(-2a)^2+2b^2=6b^2\implies b=\pm1/\sqrt6.
$$
Substituting backwards and taking into accoun the symmetric role of the variables $a,b,c$, we see that the critical points are cyclic permutations of $(a,b,c)=(2,-1,-1)/\sqrt6$ and $(a,b,c)=(-2,1,1)/\sqrt6.$ Note: we have not verified that all these points actually are critical! We could do that easily, but let's skip it, because for the purpose of finding the extremal values it is enough to have a small set of candidate points.
It is easy to verify (compute the values of $A$) that the former set of three points are the maxima, and the latter set the minima.
The answer to the original question is thus that maximum value of $A$ is
$$
A(2/\sqrt6,-1/\sqrt6,-1/\sqrt6)=\frac{2^{333}-2}{6^{166}\sqrt6}.
$$
A: 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: $a^2 b^2 c^2 \le \frac{1}{54}$
A: $a+b+c=0$, 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 $x=y,xy=\dfrac{1}{6} \to x=y=\dfrac{1}{\sqrt{6}} \to b=c=-\dfrac{1}{\sqrt{6}}, a= \dfrac{2}{\sqrt{6}}$
