let $AP \cap \omega=A_{1}$ and $PD \bot BC$ ($D$ is on $BC$).it's easy to show that $DEF$ and $A_{1}B_{1}C_{1}$ are similar so let circumradii of $DEF$ be $R'$ then $\frac{EF}{B_{1}C_{1}}=\frac{R'}{R}$ .so it's suffices to show that $R'\geq r$.let $O'$ to be the circumcenter of $DEF$ and $E' , F' , D'$ be the feet of perpendiculars from $O'$ to $AC , AB , BC$.it's obvious that $R' \geq max\left \{ O'E',O'F',O'D' \right \}=M$ so we will prove that $M\geq r$ if not then $r.p>M.p\geq S=r.p$ but this is contradiction so we are done.($p=\frac{AB+BC+AC}{2}$)

$P$ is a point in the $\vartriangle ABC$, $\omega $ is the circumcircle of $\vartriangle ABC $. $B P \cap \omega = \{ {B, B_1}\}$, $C P \cap \omega =\{ {C, C_1} \}$, $P E \perp A C$ ,$PF \perp AB$. The radius of the inscribed circle and circumcircle of $\Delta ABC $ is $r,R$ respectively. Prove that $$\frac{{EF}}{{{B_1}{C_1}}} \ge\frac{r}{R}.$$