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let $x,y,z\ge 0$,and such $$x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$$

Find the maximum of the $C$

witout loss of we assume that $$x+y+z=1$$ I think $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)=1-3(yz+xz+xy)$$

then $$(x-y)(y-z)(x-z)=?$$ so I can't，It is said $$C_{max}=\sqrt{9+6\sqrt{3}}$$

math110
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2 Answers2

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This is the standard technique for this. This one is a bit tricky, but anyways. Let WLOG $x\le y\le z$ and let $y=a+x,z=b+x$. Now plug this into $$(x+y+z)\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)\ge C|(x-y)(y-z)(z-x)|$$ $$\iff(3x+a+b)(a^2-ab+b^2)\ge Cab(a-b)\quad \text{since } a\ge b$$ Now this is linear in $x$. Hence we may take $x=0$ to achieve the minimum of LHS. This gives $$(a^3+b^3)\ge Cab(a-b)$$ We take $a=tb$ with $t\ge 1$. Then $\displaystyle C_{\max}=\min_{t\ge 1}\dfrac{t^3+1}{t(t-1)}$ Which I believe is the one mentioned above. Is this working?

shadow10
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  • This seems to work but it might be easier to parametrize $(x,y)$ as $\frac{x+y}{2}( 1+t,1-t)$ with $t \in [0,1]$. The resulting quartic equation for the minimum is easier to solve and allow one to write down the $C_{max}$ in closed form (matching the $\sqrt{9+6\sqrt{3}}$ in question.) – achille hui Jul 26 '14 at 17:40
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    Your notations are slightly confusing. For example, you wrote : let $a={\sf min}(a,b,c)$ which looks like a circular definition, of what $a$ is in terms of $a$ – Ewan Delanoy Jul 26 '14 at 17:40
  • I don't understand how $a=\min (a,b,c)$ is confusing. Suppose $(a,b,c)=(1,2,3)$ then $1=\min (1,2,3)$. And I had done the problem by replacing the variables $x,y,z$ by $a,b,c$ mistakenly.I am editing it now. I hope the confusion is a bit relieved, otherwise tell me so. Because I do hope this is a correct solution. – shadow10 Jul 27 '14 at 04:52
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    It's a little confusing notationally, but moreover unnecessary: just say that $x \leq y \leq z$ at the start, which you can do without loss of generality ($x$, $y$, and $z$ just being labels for numbers). – Semiclassical Jul 27 '14 at 05:00
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    The issue, I think, is that rather than saying that WLOG we pick $a$ to be the minimum of $a$, $b$, and $c$, it looks like you're saying "take the minimum of $a$, $b$, and $c$, and then assign that value to a new $a$". – syusim Jul 27 '14 at 05:01
  • OH! I see it now. But the assumption doesn't harm generality, since the inequality given is symmetric. And I am editing it too assume the ordering firsthand, thanks. – shadow10 Jul 27 '14 at 05:05
  • Thanks, @r9m :D – shadow10 Jul 29 '14 at 15:17
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this might be helpful

$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$

user2838619
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