<s>[INST] If $ x^2+y^2+z^2 =1$ for $x,y,z \in \mathbb{R}$, then find maximum value of $ x^3+y^3+z^3-3xyz $. If $ x^2+y^2+z^2 =1$, for $x,y,z \in \mathbb{R}$, what is the maximum of 
$ x^3+y^3+z^3-3xyz $ ?
I factorize it... Then put the maximum values of $x+y+z$ and min value of $xy+yz+zx$...
But it is wrong as they don't hold simultaneously!
Also can it be solved using partial differentiation ?
And plz provide a solution without it.!
 [/INST] because we find the maximum,so let $A>0$
let $$x^3+y^3+z^3-3xyz=A>0\Longrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=A$$
so
$$x^2+y^2+z^2=\dfrac{A}{x+y+z}+xy+yz+xz=\dfrac{A}{x+y+z}+\dfrac{1}{2}[(x+y+z)^2-x^2-y^2-z^2]$$
By AM-GM inequality
\begin{align*}\Longrightarrow \dfrac{3}{2}(x^2+y^2+z^2)&=\dfrac{A}{x+y+z}+\dfrac{1}{2}(x+y+z)^2\\
&=\dfrac{A}{2(x+y+z)}+\dfrac{A}{2(x+y+z)}+\dfrac{1}{2}(x+y+z)^2\\
&\ge \dfrac{3}{2}\sqrt[3]{A^2}
\end{align*}
so
$$A\le 1$$
</s>