Here it is.

sororak wrote: Let $x,y,z\in\mathbb{R}$ and $xyz=-1$. Prove that: \[ x^4+y^4+z^4+3(x+y+z)\geq\frac{x^2}{y}+\frac{x^2}{z}+\frac{y^2}{x}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{z^2}{y}. \] subs $z=-\frac{1}{xy}$ to the inequality, we get $LHS - RHS = \frac{{\left( {(1 + xy^2 )^2 + (1 + x^2 y)^2 + x^2 y^2 (x - y)^2 } \right)(1 - xy^2 - x^2 y)^2 }}{{2x^4 y^4 }}.$

AS xyz＝－1，So x^4+y^4+z^4+3(x+y+z)－(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y) ＝x^4+y^4+z^4－3(x+y+z)xyz－x^2(y+z)yz－y^2(z+x)zx－z^2(x+y)xy ＝ x^4+y^4+z^4－3(x+y+z)xyz－x^3(y+z)xyz－y^3(z+x)zxy－z^3(x+y)xyz ＝ x^4+y^4+z^4－3(x+y+z)xyz+x^3(y+z)+y^3(z+x)+z^3(x+y) ＝ x^4+x^3(y+z)+y^4+y^3(z+x)+z^4+z^3(x+y)－3(x+y+z)xyz ＝(x+y+z)(x^3+y^3+z^3－3xyz)＝(x+y+z)2(x^2+y^2+z^2－xy－yz－zx) ＝1/2(x+y+z)^2[(x－y)^2+(y－z)^2+(z－x)^2]≥0.

Multiplying the RHS by $x^2y^2z^2$ we get \begin{align*} x^4+y^4+z^4+3(x+y+z) &\ge x^2y^2z^2(\frac{x^2}{y}+\frac{x^2}{z}+\frac{y^2}{x}+\frac{y^2}{z}+\frac{z^2}{x}+\frac{z^2}{y}) \\ &\ge x^4yz^2+x^4y^2z+xy^4z^2+x^2y^4z+xy^2z^4+x^2yz^4 \\ &\ge x^4yz(z+y)+xy^4z(z+x)+xyz^4(y+x) \\ &\ge -x^3(y+z)-y^3(x+z)-z^3(x+y) \end{align*}Moving the terms to the LHS we get \[ x^4+y^4+z^4+3(x+y+z)+x^3(y+z)+y^3(x+z)+z^3(x+y)=x^3(x+y+z)+y^3(x+y+z)+z^3(x+y+z)+3(x+y+z)=(x^3+y^3+z^3+3)(x+y+z) \]It remains to prove that \[ (x^3+y^3+z^3+3)(x+y+z)=(x^3+y^3+z^3-3xyz)(x+y+z) \ge 0 \]Factorizing we get \[ (x+y+z)^2(x^2+y^2+z^2-xy-xz-yz) \ge 0. \]Then by AM-GM or rearrangement we get $x^2+y^2+z^2 \ge xy+xz+yz$ and we are done.

$LHS=(x^2+y^2+z^2)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=-(x^2+y^2+z^2)(xy+yz+zx)$ rest is case work for 1 negative 2 positive 3 negative and rest is obvious