Prove when $abc=1$: $ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$ 
Question: Prove the following inequality which holds for all positive reals $a$, $b$ and $c$ such that $abc=1$:
   $$ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$$

My thoughts were turning the right hand side to $abc$ as $abc=1$ however I think this will make the proving even more harder. I also  attempted applying the Cauchy-Schwarz inequality and Hölder's inequality but to no avail.
Could someone please show me how to prove it using the inequalities above or another method.

Substitute $$\frac{x}{y} \ \textrm{for}\ \ a,\quad \frac{y}{z} \ \textrm{for}\ \  b\quad \textrm{and} \quad \frac{z}{x} \ \textrm{for} \ \ c$$ This leaves us with the following and the condition $abc = 1$ is eliminated.
$$
\frac{\frac{x}{y}} {2 + \frac{y}{x}} +
\frac{\frac{y}{z}} {2 + \frac{z}{y}} +
\frac{\frac{z}{x}} {2 + \frac{x}{z}} =
\frac{x^2}{y^2 + 2xy} +
\frac{y^2}{z^2 + 2yz} +
\frac{z^2}{x^2 + 2zx} 
$$
By Titu's Lemma,
$$
\frac{x^2}{y^2 + 2xy} +
\frac{y^2}{z^2 + 2yz} +
\frac{z^2}{x^2 + 2zx} \geqslant
\frac{(x +  y + z)^2}{\sum_{cyc}{x^2} + 2xy} = \frac{(x +  y + z)^2}{(x +  y + z)^2} = 1
$$
Hence Proved.
