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Let $\{x,y,z\}\subset[0,+\infty)$,and $x+y+z=6$. Show that: $$xyz(x-y)(x-z)(y-z)\le 27$$

I tried AM -GM but without success. $$xyz\le\left(\dfrac{x+y+z}{3}\right)^3=8$$ maybe $$(x-y)(x-z)(y-z)\le \dfrac{27}{8}$$ it doesn't always true。

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  • $\begingroup$ Are you sure it is $x - z$ and not $z - x$ ? $\endgroup$ – DanielV Jul 27 '17 at 3:41
  • $\begingroup$ @DanielV,It's $x-z$ not $z-x$ $\endgroup$ – wightahtl Jul 27 '17 at 4:26
  • $\begingroup$ @TaisukeYasuda, so it's not $(z-x)$ $\endgroup$ – wightahtl Jul 27 '17 at 4:27
  • $\begingroup$ AM-GM inequality holds for nonnegative numbers. If $x \geq y \geq z$ then $x-y,x-z,y-z \geq 0$. If $x-y,y-z > 0$ then $z-x < 0$. $\endgroup$ – Zach Teitler Jul 27 '17 at 5:24
  • $\begingroup$ (1) It seems to be possible to reach 27 with values $x \approx 3.878963$ and $y \approx 1.654179$ and $z \approx 0.4668588$. Also (2) AM-GM only achieves equality when the sums and products are equal (that is, $\frac{a + b}{2} \ge \sqrt{ab}$ but $\frac{a + b}{2} = \sqrt{ab}$ only when $a=b$). $$ $$ So if you wish to apply AM-GM on terms $a_0 + a_1 + a_2 + \dots$ then you are going to have to pick the $a$ terms such that they are all equal when $x,y,z$ have the values given in (1), because it is possible to reach the value of 27 nothing less will be sufficient. $\endgroup$ – DanielV Jul 29 '17 at 11:30
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We can assume that $(x-y)(x-z)(y-z)\geq0$.

Let $x+y+z=3u$, $xy+xz+yz=3v^2$, $xyz=w^3$ and $u=tw$.

Hence, we need to prove that $$(x+y+z)^6\geq1728xyz(x-y)(x-z)(y-z)$$ or $$27u^6\geq64w^3(x-y)(x-z)(y-z)$$ or $$729u^{12}\geq4096w^6(x-y)^2(x-z)^2(y-z)^2$$ or $$27u^{12}\geq4096w^6(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $f(v^2)\geq0$, where $$f(v^2)=27u^{12}-4096w^6(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6).$$ But $f'(v^2)=24576w^6(2v^4-u^2v^2-uw^3)$, which says that $(v^2)_{min}=\frac{u^2+\sqrt{u^4+8uw^3}}{4}$.

Thus, it's enough to prove that $$f\left(\frac{u^2+\sqrt{u^4+8uw^3}}{4}\right)\geq0$$ or $$27t^{12}\geq4096\left(3t^2\left(\tfrac{t^2+\sqrt{t^4+8t}}{4}\right)^2-4\left(\tfrac{t^2+\sqrt{t^4+8t}}{4}\right)^3-4t^3+6t\left(\tfrac{t^2+\sqrt{t^4+8t}}{4}\right)-1\right)$$ or $$(t^3+8)\left(27t^9-216t^6+1216t^3+512-512\sqrt{t^3(t^3+8)}\right)\geq0.$$ Let $t^3=a$.

Hence, we need to prove that $$27a^3-216a^2+1216a+512\geq512\sqrt{a(a+8)}$$ or $$(3a-8)^2(81a^4-864a^3+7296a^2-10240a+4096)\geq0,$$ which is obvious.

Done!

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  • $\begingroup$ Nice,Have nice methods? it is said can use AM-GM inequality to solve it $\endgroup$ – wightahtl Jul 27 '17 at 4:46
  • $\begingroup$ @wightahtl I'll try to find it. $\endgroup$ – Michael Rozenberg Jul 27 '17 at 4:47
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Note that $xyz(x-y)(x-z)(y-z)$ is cyclic. WLOG, assume that $z = \min(x, y, z)$.

We only need to prove the case when $(x-y)(x-z)(y-z) > 0$. In other words, we only need to prove the case when $x > y > z$.

Let $$A = 2 + 2\cos \frac{\pi}{9}, \quad B = 2 - 2\cos \frac{4\pi}{9}, \quad C = 2 - 2\cos \frac{2\pi}{9}.$$ Clearly $A > B > C > 0$. Using AM-GM, we have \begin{align} &xyz(x-y)(x-z)(y-z)\\ =\ & \frac{x}{A}\, \frac{y}{B}\, \frac{z}{C}\, \frac{x-y}{A-B}\, \frac{x-z}{A-C}\, \frac{y-z}{B-C} \cdot ABC(A-B)(A-C)(B-C)\\ \le\ & \frac{1}{6^6}\Big(\frac{x}{A}+ \frac{y}{B}+ \frac{z}{C}+ \frac{x-y}{A-B}+ \frac{x-z}{A-C}+ \frac{y-z}{B-C}\Big)^6 \cdot ABC(A-B)(A-C)(B-C) \\ =\ & \frac{1}{6^6}(ax + by + cz)^6 \cdot ABC(A-B)(A-C)(B-C) \end{align} where $$a = \frac{1}{A} + \frac{1}{A-B} + \frac{1}{A-C}, \quad b = \frac{1}{B} - \frac{1}{A-B} + \frac{1}{B-C}, \quad c = \frac{1}{C} - \frac{1}{A-C} - \frac{1}{B-C}.$$ We can prove that $a = b = c = 1$ and $ABC(A-B)(A-C)(B-C) = 27$. We are done.

Remark: The proof of $a = b = c = 1$ and $ABC(A-B)(A-C)(B-C) = 27$ may be not simple. One method is to use discriminant and resultant by noting that $A, B, C$ are the three distinct real roots of $u^3-6u^2+9u-3 = 0$. Omitted.

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