show that $xyz(x-y)(x-z)(y-z)\le 27$

Question

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。

Answer 1

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!