the_creater wrote: $x+y+z=0$ $x^2+y^2+z^2=6$ $xy+yz+zx=-3$ by A.M.$\ge$G.M., we get $(x-y)^2+(y-z)^2+(z-x)^2\ge3{(x-y)(y-z)(z-x)}^{\frac{2}{3}}$ $|(x-y)(y-z)(z-x)|\le(\frac{(x-y)^2+(y-z)^2+(z-x)^2}{3})^{\frac{3}{2}}$ $|(x-y)(y-z)(z-x)|\le(\frac{2*6-2*3}{3})^{\frac{3}{2}}=2^{\frac{3}{2}}=2\sqrt{2}$ Are you sure that the equality holds?

crazyfehmy wrote: Let $x, y, z$ be real numbers satisfying $x+y+z=0$ and $x^2+y^2+z^2=6$. Find the maximum value of \[ |(x-y)(y-z)(z-x) | \] AmGm gives $6\sqrt3 $ as maximmum at $(\sqrt3,-\sqrt3,0)$

the_creater wrote: $x+y+z=0$ $x^2+y^2+z^2=6$ $xy+yz+zx=-3$ by A.M.$\ge$G.M., we get $(x-y)^2+(y-z)^2+(z-x)^2\ge3{(x-y)(y-z)(z-x)}^{\frac{2}{3}}$ $|(x-y)(y-z)(z-x)|\le(\frac{(x-y)^2+(y-z)^2+(z-x)^2}{3})^{\frac{3}{2}}$ $|(x-y)(y-z)(z-x)|\le(\frac{2*6-2*3}{3})^{\frac{3}{2}}=2^{\frac{3}{2}}=2\sqrt{2}$ There are two issues with this solution. You have a calculation error when evaluating $\sum(x-y)^2$, since $xy+yz+zx=-3$, not positive $3$. Even if your calculations were correct, when does equality hold? According to your solution, it does when $(x-y)^2=(y-z)^2=(z-x)^2$, but this requires two of $x,y,z$ to be equal, making the product $0$. Anyway, here's a solution that hopefully works. From $x+y+z=0$ we know that $z=-(x+y)$. Substituting this into the second equation gives \[x^2+y^2+(x+y)^2=6\implies x^2+xy+y^2=3.\] Similarly, $y^2+yz+z^2=3$ and $z^2+zx+x^2=3$. Now it suffices to maximize the square of our product, which is \[\prod(x-y)^2=\prod(x^2-2xy+y^2)=\prod(3-3xy)=27\prod(1-xy).\] Expansion gives \begin{align*}(1-xy)(1-yz)(1-zx)&=1-xy-yz-zx+x^2yz+xy^2z+xyz^2-(xyz)^2\\&=1-(-3)+0-(xyz)^2=4-(xyz)^2.\end{align*} By the Trivial Inequality, $(xyz)^2\geq 0$, so \[\prod(x-y)^2=27\prod(1-xy)\leq 27\times 4=108,\] which means that $|(x-y)(y-z)(z-x)|\leq\boxed{6\sqrt 3}$. Equality holds when, say, $(x,y,z)=(0,\sqrt 3,-\sqrt 3)$.

my solution is similar djmathman only when finishing i have used $(a^2+b^2+(a+b)^2)^3 \ge 54a^2b^2(a+b)^2,a=x-y,b=y-z$

Let $a=x-y,\ b=y-z,\ c=z-x$, $a+b+c=0$ and $(x-y)^2+(y-z)^2+(z-x)^2+(x+y+z)^2=3(x^2+y^2+z^2)=18$ $\Longleftrightarrow a^2+b^2+c^2=18.$ $\therefore ab+bc+ca=\frac12\{(a+b+c)^2-(a^2+b^2+c^2)\}=-9$, therefore, let $abc=k$, by Vieta, $a,\ b,\ c$ are the roots of the cubic equation in $t$, $t^3-9t-k=0\Longleftrightarrow k=t^3-9t:=f(t)$, $a,\ b,\ c$ are the $t$ coordinates of the points of intersection of the cubic function $y=f(t)$ and the line $y=k$, therefore, we are to find the condition for which these graphs have some point of intersection. From $f'(t)=3(t+\sqrt{3})(t-\sqrt{3})$, we have the local maximum $f(-\sqrt{3})=6\sqrt{3}$ and local minimum $f(\sqrt{3})=-6\sqrt{3}.$ Consequently, the condition stated above is $-6\sqrt{3}\leq k=abc\leq 6\sqrt{3}$, yielding $|abc|\leq 6\sqrt{3}.$ Equality holds when the cubic function touches the line $y=\pm 6\sqrt{3}$, we have $\{a,\ b,\ c\}=\{-\sqrt{3},\ -\sqrt{3},\ 2\sqrt{3}\},\ \{-2\sqrt{3},\ \sqrt{3},\ \sqrt{3}\}.$

WLOG $x\ge y\ge z$, hence $x-z\le \sqrt{2(x^2+y^2+z^2)}=2\sqrt{3}$, $ |(x-y)(y-z)(z-x) |=(x-y)(y-z)(x-z)$ $\le (\frac{x-y+y-z}{2})^2(x-z)=\frac{1}{4}(x-z)^3 \le 6\sqrt{3} $. when $x=\sqrt{3},y=0,z=-\sqrt{3}$,the maximum of $ |(x-y)(y-z)(z-x) | $ is $6\sqrt{3} $.

This problem is equivalent to Let $x, y, z$ be real numbers satisfying $(y-z)^2+(z-x)^2+(x-y)^2=6$. Find the maximum value of $ |(y+z-2x)(z+x-2y)(x+y-2z) |.$

Is $(y-z)^2+(z-x)^2+(x-y)^2=6?$

Yeah. $x\rightarrow y-z$ , ...

How about the maximum value of $|x-y|^3+|y-z|^3+|z-x|^3?$

i think global derivative $ \rightarrow $ CD3 also killed!

By using the fact $z=-a-b$ its very easy to solve the question.

Could you show us your solution? My solution is using differentiation, that's why I would like to see the solution without using differentiation.

In fact dear kunny, my solution is same as yours but its a bit easier to apply my solution , thats all.

For which problem?

kunny wrote: How about the maximum value of $|x-y|^3+|y-z|^3+|z-x|^3?$ Hello sir is there any side condition?We will denote this expression as $f(x,y,z)$ and see $f(x,y,z)=f(x+d,y+d,z+d)$ So we can easily assume $\text{min}(x,y,z)=0$ say $z=0$.Now the expression is : $|x-y|^3+|x|^3+|y|^3$ now this is going to be unbounded if we let $x\rightarrow +\infty$ I don't know if I am wrong but please correct me if I am...Thanks in advance.

Sorry, I had less explanation. I intend to post the problem of finding the maximum of $|x-y|^3+|y-z|^3+|z-x|^3$ in constrain to the condition of the original problem.

crazyfehmy wrote: Let $x, y, z$ be real numbers satisfying $x+y+z=0$ and $x^2+y^2+z^2=6$. Find the maximum value of \[ |(x-y)(y-z)(z-x) | \] http://www.artofproblemsolving.com/community/c6h1274941p6680428: Let $a,b,c$ be real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=2$. Determine the maximum value of $$|a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)|$$

Form old my problem Quote: Let $x, y, z$ be real numbers satisfying $x+y+z=0$ and $x^2+y^2+z^2=6t^2, \; (t \geqslant 0)$ then \[ |(x-y)(y-z)(z-x) | \leqslant 6\sqrt3t^3.\]

Similar to other solutions. Answer:$6\sqrt3$ and the equality holds when $(\sqrt3,-\sqrt3,0)$ Proof: $z=-x-y$ so $x^2+y^2+(x+y)^2=6$ which gives us that $x^2+xy+y^2=3$ \[(x-y)(y-z)(z-x)| \stackrel{?}{\leq} 6\sqrt3 \]\[|(x-y)(2x+y)(x+2y) \stackrel{?}{\leq} 6\sqrt3\]\[(x-y)^2(2x+y)^2(x+2y)^2\stackrel{?}{\leq} 108\]$x^2+xy+y^2=3$ so $(2x+y)^2+3y^2=12=(x+2y)^2+3x^2$ By writing $12-3x^2$ and $12-3y^2$ instead of $(x+2y)^2$ and $(2x+y)^2$, \[(x-y)^2(12-3x^2)(12-3y^2) \stackrel{?}{\leq} 108\]\[(x-y)^2(4-x^2)(4-y^2)\stackrel{?}{\leq} 12\]Let $x-y=u,xy=v$ We have $u^2+3v=3$ \[u^2(16-4(3-v)+v^2)\stackrel{?}{\leq} 12\]\[u^2(v^2+4v+4)\stackrel{?}{\leq} 12\]\[u^2(v+2)^2\stackrel{?}{\leq} 12\]We know that $6\geq u^2+2v=u^2+2(1-\frac{u^2}{3})=\frac{1}{3}u^2+2\implies 12\geq u^2$ which gives us that $3=u^2+3v\leq 12+3v\implies v\geq -3$ $u^2=3-3v$ so \[u^2(v+2)^2=(3-3v)(v^2+4v+4)\stackrel{?}{\leq} 12\]\[(1-v)(v^2+4v+4)\stackrel{?}{\leq} 4\]\[-v^3-3v^2+4\stackrel{?}{\leq} 4\]\[v^2(v+3)\stackrel{?}{\geq}0\]which is true.