show that inequality $(x-3)^4+(y-3)^4+(z-3)^4\ge 193$

Question

Let $x,y,z\in R$,and such $$xy+yz+xz=-1$$ show that $$(x-3)^4+(y-3)^4+(z-3)^4\ge 193$$

it seem use Cauchy-Schwarz inequality to solve it?But I try sometime can't get this answer,even now I can't find this equality when $=$?

Answer 1

Mathematica is not happy (Unless I'm blind and missed an obvious typo). $$ x := -\frac{485}{2048} \\ y := -\frac{5}{2048} \\ z := \frac{4196729}{1003520} $$ Then $xy + yz + zx = -1$ (See WA). However $$ (x-3)^4 + (y-3)^4 + (z-3)^4 = 192.98312371 < 193 $$ See WA.

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Also it seems that Mathematica FindInstance cannot disprove your statement in $\mathbb{Z}$ so it might be true in this set.

Answer 2

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

Hence, the expression $\sum\limits_{cyc}(x-3)^4$ is a linear expression of $w^3$,

which says that it gets a minimal value for an extremal value of $w^3$,

which happens for equality case of two variables.

Let $y=x$. Thus, $z=-\frac{1+x^2}{2x}$ and $$\min_{xy+xz+yz=-1}\sum_{cyc}(x-3)^4=\min\left(2(x-3)^4+\left(-\frac{1+x^2}{2x}-3\right)^4\right)=191.779...,$$ which occurs for $x_1=-0.12...$, where $x_1$ is a negative root of the following equation. $$11x^6-90x^5+321x^4-180x^3-107x^2-18x-1=0.$$