$LHS+4=\left(\frac{(a-1)(c+1)}{1+bc+c}+1\right) + \left(\frac{(b-1)(d+1)}{1+cd+d}+1\right) + \left(\frac{(c-1)(a+1)}{1+da+a} +1\right)+ \left(\frac{(d-1)(b+1)}{1+ab+b}+1\right)$ $=\frac{a+bc+ac}{1+bc+c}+\frac{b+cd+bd}{1+cd+d}+\frac{c+da+ca}{1+da+a}+\frac{d+ab+bd}{1+ab+b}$ $=\frac{(a+bc+ac)(\frac{1}{a}+bc+\frac{c}{a})}{(1+bc+c)(\frac{1}{a}+bc+\frac{c}{a})}+\frac{(b+cd+bd)(\frac{1}{b}+cd+\frac{d}{b})}{(1+cd+d)(\frac{1}{b}+cd+\frac{d}{b})}+\frac{(c+da+ca)(\frac{1}{c}+da+\frac{a}{c})}{(1+da+a)(\frac{1}{c}+da+\frac{a}{c})}+\frac{(d+ab+bd)(\frac{1}{d}+ab+\frac{b}{d})}{(1+ab+b)(\frac{1}{d}+ab+\frac{b}{d})}$ $\ge\frac{(1+bc+c)^2}{(1+bc+c)(\frac{1}{a}+bc+\frac{c}{a})}+\frac{(1+cd+d)^2}{(1+cd+d)(\frac{1}{b}+cd+\frac{d}{b})}+\frac{(1+da+a)^2}{(1+da+a)(\frac{1}{c}+da+\frac{a}{c})}+\frac{(1+ab+b)^2}{(1+ab+b)(\frac{1}{d}+ab+\frac{b}{d})}$ $=\frac{a(1+bc+c)}{(1+cd+d)}+\frac{b(1+cd+d)}{(1+da+a)}+\frac{c(1+da+a)}{(1+ab+b)}+\frac{d(1+ab+b)}{(1+bc+c)}\ge 4=RHS+4$ PS : oops actually my solution is just the same with the above solution, only without the subs

Let $a, b, c$ be positive real numbers such that $abc= 1$. Prove that \[\frac{(a-1)(c+1)}{1+bc+c} + \frac{(b-1)(a+1)}{1+ca+a} + \frac{(c-1)(b+1)}{1+ab+b} \geq 0.\]

$abcd=1 \implies$ we can make substitution: $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{t}, d=\frac{t}{x}$. Then our inequality becomes to: \[\frac{(x-y)(z+t)}{y(y+z+t)}+\frac{(y-z)(t+x)}{z(x+z+t)}+\frac{(z-t)(x+y)}{t(x+y+t)}+\frac{(t-x)(y+z)}{x(x+y+z)}\ge 0\] It is equivalent to \begin{align*} \left(\frac{(x-y)(z+t)}{y(y+z+t)}+1\right)+\left(\frac{(y-z)(t+x)}{z(x+z+t)}+1\right)+\left(\frac{(z-t)(x+y)}{t(x+y+t)}+1\right)+\left(\frac{(t-x)(y+z)}{x(x+y+z)}+1\right)\ge 4. \end{align*} or \[\frac{y^{2}+xz+xt}{y(y+z+t)}+\frac{z^{2}+yt+xy}{z(x+z+t)}+\frac{t^{2}+zx+yz}{t(x+y+t)}+\frac{x^{2}+ty+tz}{x(x+y+z)}\ge 4\] From Cauchy-Schwarz inequality we have: $(y^{2}+xz+xt)(1+\frac{z}{x}+\frac{t}{x})\ge (y+z+t)^{2}$. This means \[\frac{y^{2}+xz+xt}{y(y+z+t)}\ge \frac{x(y+z+t)}{y(x+z+t)}.\] Similarly, we have \begin{align*} \frac{z^{2}+yt+xy}{z(x+z+t)}\ge \frac{y(x+z+t)}{z(x+y+t)}, \quad \frac{t^{2}+zx+yz}{t(x+y+t)}\ge \frac{z(x+y+t)}{t(x+y+z)}, \quad \frac{x^{2}+ty+tz}{x(x+y+z)}\ge \frac{t(x+y+z)}{x(y+z+t)}. \end{align*} So, we need to prove that \[\frac{x(y+z+t)}{y(x+z+t)}+\frac{y(x+z+t)}{z(x+y+t)}+\frac{z(x+y+t)}{t(x+y+z)}+\frac{t(x+y+z)}{x(y+z+t)}\ge 4.\] It holds from $AM-GM$. Because product of fractions is equal to 1. Nice problem!

I have the same solution but I used Cauchy-Shwartz in another form: $ \frac{xt+xz+y^{2}}{y+t+z}= \frac{t^{2}}{ \frac{yt}{x}+ \frac{t^{2}}{x}+ \frac{zt}{x}}+ \frac{z^{2}}{ \frac{yz}{x} + \frac{tz}{x} + \frac{z^{2}}{x}} + \frac{y^{2}}{y+t+z} \geq \frac{x(y+t+z)^{2}}{(y+t+z)(x+t+z)} = \frac{x(y+t+z)}{x+t+z}$

sqing wrote: Let $a, b, c$ be positive real numbers such that $abc= 1$. Prove that \[\frac{(a-1)(c+1)}{1+bc+c} + \frac{(b-1)(a+1)}{1+ca+a} + \frac{(c-1)(b+1)}{1+ab+b} \geq 0.\] $\frac{(a-1)(c+1)}{1+bc+c} + \frac{(b-1)(a+1)}{1+ca+a} + \frac{(c-1)(b+1)}{1+ab+b} \geq 0.$ $\Leftrightarrow \frac{a+b}{1+bc+c} + \frac{b+c}{1+ca+a} + \frac{c+a}{1+ab+b} \geq2.$ $\Leftrightarrow (\frac{1}{1+bc+c} + \frac{1}{1+ca+a} + \frac{1}{1+ab+b} )(a+b+c) \geq3.$ $\Leftrightarrow a+b+c \geq3.$ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2920865 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=556256&p=3248160#p3248160

matrix41 wrote: $LHS+4=\left(\frac{(a-1)(c+1)}{1+bc+c}+1\right) + \left(\frac{(b-1)(d+1)}{1+cd+d}+1\right) + \left(\frac{(c-1)(a+1)}{1+da+a} +1\right)+ \left(\frac{(d-1)(b+1)}{1+ab+b}+1\right)$ $=\frac{a+bc+ac}{1+bc+c}+\frac{b+cd+bd}{1+cd+d}+\frac{c+da+ca}{1+da+a}+\frac{d+ab+bd}{1+ab+b}$ $=\frac{(a+bc+ac)(\frac{1}{a}+bc+\frac{c}{a})}{(1+bc+c)(\frac{1}{a}+bc+\frac{c}{a})}+\frac{(b+cd+bd)(\frac{1}{b}+cd+\frac{d}{b})}{(1+cd+d)(\frac{1}{b}+cd+\frac{d}{b})}+\frac{(c+da+ca)(\frac{1}{c}+da+\frac{a}{c})}{(1+da+a)(\frac{1}{c}+da+\frac{a}{c})}+\frac{(d+ab+bd)(\frac{1}{d}+ab+\frac{b}{d})}{(1+ab+b)(\frac{1}{d}+ab+\frac{b}{d})}$ $\ge\frac{(1+bc+c)^2}{(1+bc+c)(\frac{1}{a}+bc+\frac{c}{a})}+\frac{(1+cd+d)^2}{(1+cd+d)(\frac{1}{b}+cd+\frac{d}{b})}+\frac{(1+da+a)^2}{(1+da+a)(\frac{1}{c}+da+\frac{a}{c})}+\frac{(1+ab+b)^2}{(1+ab+b)(\frac{1}{d}+ab+\frac{b}{d})}$ $=\frac{a(1+bc+c)}{(1+cd+d)}+\frac{b(1+cd+d)}{(1+da+a)}+\frac{c(1+da+a)}{(1+ab+b)}+\frac{d(1+ab+b)}{(1+bc+c)}\ge 4=RHS+4$ PS : oops actually my solution is just the same with the above solution, only without the subs I think it is $=\dfrac{ad(1+bc+c)}{(1+cd+d)}+\dfrac{ba(1+cd+d)}{(1+da+a)}+\dfrac{cb(1+da+a)}{(1+ab+b)}+\dfrac{dc(1+ab+b)}{(1+bc+c)}\ge 4 $ (By Cô-si and $abcd=1$)

Essentially as the previous solutions, but with a (in my opinion) more intuitive use of the Engel form of Cauchy-Schwarz. Firstly, to get rid of $abcd=1$ we use the standard substitution $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{t}$, $d=\frac{t}{x}$ and thus require $\sum_{x,y,z,t} \frac{(x-y)(z+t)}{y(y+z+t)}\geq 0$. To get rid of negative terms (as most well known applicable inequalities are for positive real numbers) we add $1$ to each of the $4$ summands, to obtain the equivalent $\sum_{x,y,z,t} \frac{xz+xt+y^2}{y(y+z+t)} \geq 4$. Now having in mind that Cauchy in Engel form is from the best inequalities to manipulate fractions, we would like to have squares in the numerators. One way to do that is to switch from $xz$ to $x^2z^2$, from $xt$ to $x^2t^2$ and keep $y^2$ as it is, but in that case things will (probably?) not work. So it would be better if we also switch from $y^2$ to $x^2y^2$. That is, rewrite $\frac{xz+xt+y^2}{y(y+z+t)}$ as $\frac{x^2z^2}{xyz(y+z+t)} + \frac{x^2t^2}{xyt(y+z+t)} + \frac{x^2y^2}{x^2y(y+z+t)}$ and then use the Engel form to obtain that the latter is at least $\frac{(xz+xt+xy)^2}{y(y+z+t)(xz+xt+x^2)} = \frac{x(y+z+t)}{y(x+z+t)}$. It remains to notice easily that $\sum_{x,y,z,t} \frac{x(y+z+t)}{y(x+z+t)} \geq 4$ by AM-GM.