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.

Another way for the first inequality.
For positives $x$, $y$ and $z$ we need to prove that:
$$\sum_{cyc}(x^3-x^2z)\geq0.$$
Indeed,
$$\sum_{cyc}(x^3-x^2z)=\sum_{cyc}(x^3-xy^2)=$$
$$=\sum_{cyc}x(x+y)(x-y)=\sum_{cyc}\left((x^2+x)(x-y)-\frac{2}{3}(x^3-y^3)\right)=$$
$$=\frac{1}{3}\sum_{cyc}(x-y)(3x^2+3xy-2x^2-2xy-2y^2)=\frac{1}{3}\sum_{cyc}(x-y)^2(x+2y)\geq0.$$
