trouble understanding proof for the inequality: $\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)} \ge \frac{3}{4}$ I have read this problem in a book discussing inequalities, and I have a few problems in understanding the proof.
The problem is:
Let $a,b,c$ be positive numbers that satisfy $abc=1$, prove that
$$\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)} \ge \frac{3}{4}$$
The solution is:
$$\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)}$$
$$=\frac{(a+1)(b+1)(c+1)-2}{(a+1)(b+1)(c+1)}=1-\frac{2}{(a+1)(b+1)(c+1)} \ge \frac{3}{4}$$
if and only if $(a+1)(b+1)(c+1) \ge 8$ and this last inequality follows immediately from the inequality $(\frac{a+1}{2})(\frac{b+1}{2})(\frac{c+1}{2}) \ge \sqrt{abc} = 1$  
I have a problem understanding the last two steps and would like someone to clarify what is actually going on.

The last inequality follows from $a+1\geq 2\sqrt{a}$ (and analogous inequalities for $b$ and $c$), which is true because $(\sqrt{a}-1)^2\geq 0$.
For the step before this, you can rearrange $$1-\frac{2}{(a+1)(b+1)(c+1)} \ge \frac{3}{4}$$ to get $$\frac{1}{4}\geq\frac{2}{(a+1)(b+1)(c+1)}$$ and $$(a+1)(b+1)(c+1)\geq 8.$$
