Let $A=\{x\in \mathbb{R}^n|\sum x_i=n/3, \sum x_i^2=n \}$
$f(x)=\sum x_i^3$
Prove that max of f on A is of the form: $x=(a,a,.....,a,b,b...,b)$ (no need to find a or b).
So with Lagrange multiplier method I used these as my constraints:
$g(x)=\sum x_i-n/3=0$
$h(x)=\sum x_i^2-n=0$
so $f_{x_i}=\lambda g_{x_i}+ \mu h_{x_i}$
we get n equations of:
$3x_i^2=\lambda+\mu2x_i$
summing them all up I deduced that $ \lambda=3-\frac{2n\mu}{3}$
however I can't find lambda and prove what's needed..
Any help?