<s>[INST] 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.
 [/INST] We have$:$
$$\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1}$$
$$={\sum \frac {2\,b{c}^{2} \left( ab-1 \right) ^{2}+b{c}^{2} \left( a-1
 \right) ^{2}}{ 3\left( ab+1 \right)  \left( bc+1 \right)  \left( ac+1
 \right) }} -\frac{f(a,b,c) }{{ 3\left( ab+1
 \right)  \left( bc+1 \right)  \left( ac+1 \right) }}\geq 0$$
where
$$f(a,b,c)= \left( abc-1 \right)  \Big[2\,abc \left( a+b+c
 \right) -6\,(ab+bc+ca)+3\,a+3\,b+3\,c-9 \Big]  =0$$
</s>