<s>[INST] 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 $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{1}{bc+bc^2+c+1}+\dfrac{1}{ac+a^2c+c+1}+\dfrac{1}{ab+ab^2+b+1}.$$
 [/INST] The inequality is equivalent to
$$4(a(1+a)+b(1+b)+c(1+c))\ge 3(1+a)(1+b)(1+c).\tag{1}$$
Since 
$$(1+a)(1+b)(1+c) = 1+ (a+b+c)+(ab+bc+ca)+abc$$
and $abc=1$, Inequality (1) is equivalent to
$$4(a^2+b^2+c^2)+(a+b+c) \ge 6+3(ab+bc+ca).$$
The last inequality follows from $a^2+b^2+c^2\ge ab+bc+ca$ and AM-GM.
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