If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I can not find a way to decompose of combine.
Also, how do you prove that $x^4+y^4+z^2 \ge xyz \sqrt{8}$?
I got $x^4+y^4+z^2\geq3\sqrt[3]{x^4*y^4*z^2}=3xy\sqrt[3]{xyz^2}.$
I do not know what to do from here.

It's wrong.
Try, $b>1$ and $bc\rightarrow-1^-$.
It's true for positive variables.
Indeed, let $a=\frac{y}{x}$ and $b=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and we need to prove that
$$\sum_{cyc}\frac{\frac{y}{x}-1}{\frac{y}{x}\cdot\frac{z}{y}+1}\geq0$$ or
$$\sum_{cyc}\frac{y-x}{x+z}\geq0$$ or
$$(y-x)(y+x)(y+z)\geq0$$ or
$$\sum_{cyc}(y-x)(y^2+xy+xz+yz)\geq0$$ or
$$\sum_{cyc}(x^3-x^2z)\geq0,$$ which is true by Rearrangement:
$$\sum_{cyc}x^3=\sum_{cyc}(x^2\cdot x)\geq\sum_{cyc}(x^2\cdot z)=\sum_{cyc}x^2z.$$
The second inequality.
By AM-GM
$$x^4+y^4+z^2=x^4+y^4+2\cdot\frac{1}{2}z^2\geq4\sqrt[4]{x^4y^4\left(\frac{1}{2}z^2\right)^2}=$$
$$=\sqrt8|xyz|\geq\sqrt8xyz.$$
