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.
 A: 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
$$x^4+y^4+z^2=x^4+y^4+2\cdot\frac{1}{2}z^2\geq4\sqrt[4]{x^4y^4\left(\frac{1}{2}z^2\right)^2}=$$
$$=\sqrt8|xyz|\geq\sqrt8xyz.$$
A: For the first inequality, using the substitution of $ a = \frac{y}{x}, b = \frac{z}{y},c = \frac{x}{z}$, (which is a good approach to nomarlize these inequalities), we WTS   (see Michaels' solution)
$$  \sum y \times \frac{1}{ (x+y+z) - y } \geq \sum x \times \frac{ 1}{(x+y+z) - y }.$$
This is just rearrangement inequality, since $\{ y\}$ and $\{ \frac{ 1}{ (x+y+z) - y } \}$ are similarly ordered.
A: 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,
$$\sum_{cyc}(x^3-x^2z)=\sum_{cyc}(x^3-xy^2)=$$
$$=\sum_{cyc}x(x+y)(x-y)=\sum_{cyc}\left((x^2+x)(x-y)-\frac{2}{3}(x^3-y^3)\right)=$$
$$=\frac{1}{3}\sum_{cyc}(x-y)(3x^2+3xy-2x^2-2xy-2y^2)=\frac{1}{3}\sum_{cyc}(x-y)^2(x+2y)\geq0.$$
A: 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
$$f(a,b,c)= \left( abc-1 \right)  \Big[2\,abc \left( a+b+c
 \right) -6\,(ab+bc+ca)+3\,a+3\,b+3\,c-9 \Big]  =0$$
