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.

If $a,b,c> 0$ and $abc=1$, then by Cauchy-Schwarz: $$\sum_{\text{cyc}}\frac{a}{2+bc}=\sum_{\text{cyc}}\frac{a^2}{2a+abc}=\sum_{\text{cyc}}\frac{a^2}{2a+1}\ge \frac{(a+b+c)^2}{2(a+b+c)+3}$$
And by AM-GM $a+b+c\ge 3\sqrt[3]{abc}=3$, so: $$(a+b+c)^2-2(a+b+c)-3=((a+b+c)-3)((a+b+c)+1)\ge 0$$
Equality holds if and only if $a=b=c=1$.
