Prove that $\frac{a}{(1+b)(1+c)}+\frac{b}{(1+a)(1+c)}+\frac{c}{(1+a)(1+b)} \geq \frac{3}{4}.$ 
Let $a,b,c$ be positive numbers that satisfy $abc = 1$, prove that $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{3}{4}.$$

Attempt
I tried doing $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)}  =  \dfrac{1}{bc(1+b)(1+c)}+\dfrac{1}{ac(1+a)(1+c)}+\dfrac{c}{bc(1+a)(1+b)}$$ then using the fact that $a^2+b^2+c^2 \geq ab+bc+ca$ but that only seemed to get me $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{1}{bc+bc^2+c+1}+\dfrac{1}{ac+a^2c+c+1}+\dfrac{1}{ab+ab^2+b+1}.$$

Ok, let's sum it:
$$\frac{(a+1)a+b(b+1)+c(c+1)}{(a+1)(b+1)(c+1)}$$, called it $(1)$
Now use Cauchy inequality:
$$(1) >= \frac{3((a+1)ab(b+1)c(c+1))^{\frac{1}{3}}}{(a+1)(b+1)(c+1)}$$, now
Derived all, what we can derive :)
And use Cauchy for $a+1,b+1,c+1$.
So continue it!
Hint: show that:
$$((a+1)(b+1)(c+1))^{\frac{2}{3}} >= 4$$
