<s>[INST] Prove that $1+b+(1+c)+1/c+1+a \ge 3$ if $a, b,$ and $c$ are positive real numbers. Let $a, b, c$ be positive real numbers. prove that
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},
$$
and that equality occurs if and only if $a = b = c = 1$.
What I tried:
$1$st part: I tried a brute force approach where I make a common denominator for all $4$ fractions, but the number get real big real fast.
 [/INST] AM-GM helps:
$$\sum_{cyc}\frac{1}{a(1+b)}=\frac{1}{1+abc}\sum_{cyc}\frac{1+abc}{a(1+b)}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+abc}{a(1+b)}+1\right)-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\right)-\frac{3}{1+abc}\geq$$
$$\geq\frac{1}{1+abc}\cdot6\sqrt[6]{\prod_{cyc}\frac{(1+a)}{a(1+b)}\prod_{cyc}\frac{b(1+c)}{1+b}}-\frac{3}{1+abc}=\frac{3}{1+abc}.$$
</s>