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.!
 A: Let $f(t)=\frac{t}{2}(3-t^2)$. It is straightforward to check that $f$ attains its maximum on the interval $[-\sqrt{3},\sqrt{3}]$ at $t=1$, with $f(1)=1$.
Now, if $x^2+y^2+z^2=1$ then
$$\eqalign{
x^3+y^3+z^3-3xyz&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\cr
&=\frac{ x+y+z }{2}\left(3(x^2+y^2+z^2)-(x+y+z)^2\right)\cr
&=f(x+y+z)
}
$$
But by Cauchy-Schwarz inequality we have
$$|x+y+z|\leq\sqrt{3}\sqrt{x^2+y^2+z^2}=\sqrt{3}$$
So,
$$
x^3+y^3+z^3-3xyz\leq f(1)=1
$$
with equality if and only if $x+y+z=1$. This corresponds to the points of the circle
of intersection of the sphere of equation $x^2+y^2+z^2=1$ with the plane of equation
$x+y+z=1$.
A: 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
$$A\le 1$$
A: There is one way without Lagrange multipliers: use spherical coordinate system and write $x,y,z$ as:
$x=\sin \theta \cos \phi$
$y=\sin \theta \sin \phi$
$z=\cos \theta$
It's equal $x^2+y^2+z^2=1$.
Now you can find maximal value using partial derivatives, because $\phi, \theta \in \mathbb{R}$
A: Yet another way, let $p = x+y+z, q = xy+yz + zx$.  Then $x^2+y^2+z^2 = p^2-2q=1$ and  $x^3+y^3+z^3-3xyz = p^3-3pq = \frac12p(3-p^2)$ .  Also $x^2+y^2+z^2 = 1 \implies p^2\le 3$.
So we want the maximum of $\frac12p(3-p^2)$, subject to $p^2 \le 3$, which easily$^\dagger$ gives $p=1$.  

$^\dagger$ Equivalently this is when $\frac14p^2(3-p^2)^2$ gets maximised, or $(p^2)\dfrac{3-p^2}2 \dfrac{3-p^2}2$.  As the last is a product of three positive terms with constant sum $3$, it gets maximised when all terms are equal, i.e. $p=1$.
A: The way to solve this is to use Lagrange multipliers to find the max of $f(x,y,z) = x^3+y^3+z^-3xyz$ given the constraint $g(x,y,z) = x^2+y^2+z^2 =1$.
Use this link for help:

http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx

So the first thing to do is to find the gradient of $f(x,y,z)$ and $g(x,y,z)$.
$$\nabla f(x,y,z) =  <3x^2 - 3yz, 3y^2 - 3xz, 3z^2-3yz>$$
$$\nabla g(x,y,z) =  <2x, 2y, 2z>$$
Then we solve the equation 
$$\nabla f(x, y, z) = \lambda \nabla g(x,y,z)$$
For all $x, y, z,$ and $\lambda$. Then you plug the values in to the original equation and figure out the maximum and minimum values.
A: The best way to approach this question is to use lagrange multipliers
Here is a link if you would like to read an example: http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx
To begin,
$$f(x, y, z) = x^3 + y^3 + z^3 - 3xyz $$
$$g(x, y, z) = x^2 + y^2 + z^2 $$
The constraint is that
$$g(x, y, z) = x^2 + y^2 + z^2 = 1 $$
Now we create our 4 equation, using the constraint and by taking the partial derivatives of the function.
$$f_x = \lambda g_x, 3x^2 = 2\lambda x ....(1)$$
$$f_y = \lambda g_y, 3y^2 = 2\lambda y ....(2)$$
$$f_z = \lambda g_z, 3z^2 = 2\lambda z ....(3)$$
$$x^2 + y^2 + z^2 = 1 ....(4)$$
Substituting equations (1), (2), and (3) into (4):
$$\frac{2\lambda x}{3} + \frac{2\lambda y}{3} + \frac{2\lambda z}{3} = 1$$
Also, notice that $$3x = 2\lambda$$
$$3y = 2\lambda$$
$$3z = 2\lambda$$
Meaning
$$x = y = z$$
So our maximum will occur when x and y and z are all equal.
Plugging this into our constraint:
$$3x^2 = 1$$ $$x = y = z = +/-\frac{\sqrt{3}}{3}$$
Notice that maximum value will only occur when x, y, and z are all positive.
Therefore, maximum value is:
$$3x^3 - 3x^3 = 0$$
