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.
 A: 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 $a=b=c=1$.
A: Note that $$\sum_{cyc}\frac{a}{bc+2}=\sum_{cyc}\frac{a^2}{2a+abc}=\sum_{cyc}\frac{a^2}{2a+1} \ge \frac{(a+b+c)^2}{2a+2b+2c+3} $$
Now one can use that $$\sum_{cyc}a \ge 3\sqrt[3]{abc}=3$$This establishes that $$(a+b+c-1)^2 \ge 4 \Leftrightarrow (a+b+c)^2 \ge 2(a+b+c)+3$$Thus, our proof is done, with equality at $a=b=c$. 
A: Let $p,q,r \in \mathbb{R}_{>0}$ such that $(p^3,q^3,r^3)=(a,b,c)$.
Then $(pqr)^3 = 1$ and hence $pqr = 1$.
Thus $\sum_{cyc} \frac{a}{2+bc} \ge 1$ iff $\sum_{cyc} \frac{p^3 (pqr)}{2 (pqr)^2 + q^3 r^3} \ge 1$ iff $\sum_{cyc} \frac{p^4}{2 p^2 q r + q^2 r^2} \ge 1$.
This technique is called homogenization, and many homogenous inequalities can be solved by expanding and using AM-GM and Schur's. I didn't try for this one though.
A: 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.
