Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ac}{1+c} \geq 3.$ 
Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1+ab}{1+a}+\dfrac{1+bc}{1+b}+\dfrac{1+ac}{1+c} \geq 3.$$

This looks symmetric, so should I prove it for just $a \leq b \leq c$ and then the other cases follow?

Another way.
Let $a=\frac{y}{x},$ $b=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Thus,  $c=\frac{x}{z}$ and  by AM-GM we obtain: $$\sum_{cyc}\frac{1+ab}{1+a}=\sum_{cyc}\frac{1+\frac{z}{x}}{1+\frac{y}{x}}=\sum_{cyc}\frac{x+z}{x+y}\geq3\sqrt[3]{\prod_{cyc}\frac{x+z}{x+y}}=3.$$
