It's equivalent to $\sum \frac{a}{1+b+ab} \ge \sum \frac{b}{1+b+ab}$ which after full expansion is just $x^2+y^2+z^2 \ge xy+yz+zx$ for $x=1+b+ab$ , $y=1+c+bc$ and $z=1+a+ac$.

more detail plz

Firstly we do : $\frac{1+b+ab+a-b}{1+b+ab}=1+\frac{a}{1+b+ab}-\frac{b}{1+b+ab}$ and for the two other. Now we need to prove : $ \frac{a}{1+b+ab}+ \frac{b}{1+c+bc}+ \frac{c}{1+a+ac} \ge \frac{b}{1+b+ab}+ \frac{c}{1+c+bc}+ \frac{a}{1+a+ac}$. We now multiply out by $(1+b+ab)(1+c+bc)(1+a+ac)$ to get : $a(1+c+bc)(1+a+ac)+b(1+b+ab)(1+a+ac)+c(1+b+ab)(1+c+bc) \ge b(1+c+bc)(1+a+ac)+c(1+b+ab)(1+a+ac)+a(1+b+ab)(1+c+bc)$. Now note that: $a(1+c+bc)(1+a+ac)=(a+ac+abc)(1+a+ac)=(1+a+ac)^2$ and two similar. For the right side we just do:$b(1+c+bc)(1+a+ac)=(1+c+bc)(1+b+ab)$, now I think my first post should be clear.

Note that $\frac{1+a+ab}{1+b+ab}+\frac{1+b+bc}{1+c+bc}+\frac{1+c+ac}{1+a+ac} =\frac{(1+a+ab)+ab(1+b+bc)+b(1+c+ac)}{1+b+ab}$ (from $abc=1$) which is equal to $\frac{1+b+ab+a+ab+ab^2+b+bc+1}{1+b+ab}=\frac{2(1+b+ab)+ab^2+a+bc}{1+b+ab}=2+\frac{a+ab^2+bc}{1+b+ab}$. Now $a+ab^2+bc \geq 1+b+ab$ which comes from taking $a=x^2, ab^2=y^2, bc=z^2$ and $x^2+y^2+z^2 \geq xy+yz+zx$. Thus, $\frac{1+a+ab}{1+b+ab}+\frac{1+b+bc}{1+c+bc}+\frac{1+c+ac}{1+a+ac}=2+\frac{a+ab^2+bc}{1+b+ab} \geq 2+\frac{1+b+ab}{1+b+ab}=3$. Proved.

Let $a$, $b$, and $c$ be real numbers such that $abc=1$. prove that $$\frac{1+a+ab}{1+b+kab}+\frac{1+b+bc}{1+c+kbc}+ \frac{1+c+ca}{1+a+kca}\geq\frac{9}{k+2} $$Where $k\geq 1.$

k=2

Ejaifeobuks wrote: Let $a$, $b$, and $c$ be real numbers such that $abc=1$. prove that $$\frac{1+a+ab}{1+b+ab}+\frac{1+b+bc}{1+c+bc}+ \frac{1+c+ac}{1+a+ac}\geq3$$

$$\iff$$$$\frac{a}{1+b+ab}+\frac{b}{1+c+bc}+ \frac{c}{1+a+ac}\geq1$$$$\iff$$$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$h

Attachments:

This reduces to trivial inequality! Use cyclic expression trick to combine expressions: $$\frac{(c+ac+1)+(1+b+bc)+(bc+bc^2+c)}{1+c+bc} \geq 3$$Kill $a$: $$b^2+1+b^2c^2 \geq b+bc+b^2c$$$$(b-1)^2+(bc-1)^2+(bc-b)^2 \geq 0$$

Here is a link to a video of the solution. https://youtu.be/u63a0HjOudo