If $abcd=1$ prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+cd}{1+c}+\frac{1+ad}{1+d} \ge 4$

Question

If $abcd=1$ prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+cd}{1+c}+\frac{1+ad}{1+d} \ge 4.$

I tried a long bruteforce method and came up with

$(1+b)(1+c)(1+a)(1-a)+(1+c)(1+d)(1+b)(1-b)+(1+a)(1+d)(1+c)(1-c)+(1+a)(1+b)(1+d)(1-d) \ge 0.$

But I am struck here. How to go further??

Answer 1

It's true for positive variables.

Indeed, let $a=\frac{y}{x},$ $b=\frac{z}{y}$ and $c=\frac{t}{z},$ where $x$, $y$, $z$ and $t$ are positives.

Thus, $d=\frac{x}{t}$ and by C-S we obtain: $$\sum_{cyc}\frac{1+ab}{1+a}=\sum_{cyc}\frac{1+\frac{z}{x}}{1+\frac{y}{x}}=\sum_{cyc}\frac{x+z}{x+y}=$$ $$=(x+z)\left(\frac{1}{x+y}+\frac{1}{z+t}\right)+(y+t)\left(\frac{1}{y+z}+\frac{1}{t+x}\right)\geq$$ $$\geq\frac{4(x+z)}{x+y+z+t}+\frac{4(y+t)}{x+y+z+t}=4.$$