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.!
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 $x+y+z=1$.
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 $$A\le 1$$
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 $\phi, \theta \in \mathbb{R}$
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. $p=1$.
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.
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:
$$3x^3 - 3x^3 = 0$$
