Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$ I tried doing it with CS-Engel to get $$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c}
$$
I thought that maybe proof that $$
\frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)}
$$ or $$
3+3 a b c \geq a+b+c+a b+b c+a c
$$, but I don't know how
 A: For positive variables by AM-GM we obtain:
$$\sum_{cyc}\frac{1}{a(1+b)}=\frac{1}{1+abc}\sum_{cyc}\frac{1+abc}{a(1+b)}=\frac{1}{1+abc}\left(\sum_{cyc}\frac{1+abc}{a(1+b)}+1-1\right)=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab+abc}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\right)-\frac{3}{1+abc}\geq$$
$$\geq\frac{6}{1+abc}\sqrt[6]{\prod_{cyc}\left(\frac{1+a}{a(1+b)}\cdot\frac{b(1+c)}{1+b}\right)}-\frac{3}{1+abc}=\frac{3}{1+abc}.$$
A: As in Michael's solution, this uses the same idea of "+1" to split up the fraction.
Written up this way, it might seem like a more natural approach.

WTS
$$ \sum \frac{ 1 + abc } { a (1+b)} \geq 3 $$
$$\sum \frac{ 1 + abc + a + ab } { a (1+b) }  \geq 6$$
$$ \sum \frac{ ab ( 1 + c)  + (1+a ) } { a (1+b) } \geq 6$$
$$ \sum \frac{ ab (1+c) } { a(1+b) } + \frac{ (1+b) } { b (1 + c) } \geq 6$$
Applying AM-GM to all 6 terms, the result follow.
From the conditions, we can easily deduce that the equality case is only $ a = b = 1$.
