It was posted in mathlink, I can't find the link therefore I will post my solution By law of sine $\frac{OD}{\sin OBD}=\frac{a}{\sin BOC}$ it isn't hard to show $\angle OBD=\frac{\pi}{2}-(\angle B-\angle C),\angle BOC=2\angle A$ when $O$ inside triangle(because triangle is acute) thus $OD=\frac{a\cos(B-C)}{\sin 2A}=\frac{2R\sin A\cos(B-C)}{2sin A\cos A}=R\frac{\cos(B-C)}{cos A}$ similarly for $B,C$ thus we need to prove $\prod\cos(B-C)\ge 8\prod\cos A$ multiply with $\prod\sin A$ we have $\prod\sin A\cos(B-C)\ge 8\prod\cos A\sin A=\prod\sin 2A$ But $\prod\sin A\cos(B-C)=\prod\sin(B+C)\cos(B-C)=\prod\frac{1}{2}(sin 2B+sin 2C)$ thus we need to prove $\frac{1}{8}\prod(\sin 2B+\sin 2C)\ge\prod\sin 2A$ but it is quite obvious by $AM-GM$

N.T.TUAN wrote: Let $O$ be the circumcenter and $R$ be the circumradius of an acute triangle $ABC$. Let $AO$ meet the circumcircle of $OBC$ again at $D$, $BO$ meet the circumcircle of $OCA$ again at $E$, and $CO$ meet the circumcircle of $OAB$ again at $F$. Show that $OD.OE.OF\geq 8R^{3}$. Proof. Note $I\in BC\cap AD$ and the reflection $A'$ of the point $A$ w.r.t. the point $O$. Then $\boxed{\mathrm{\ the\ points\ A\ ,\ A'\ are\ harmonical\ conjugate\ w.r.t.\ the\ points\ I\ ,\ D\ }}$. Let $\rho$ be the circumradius of the triangle $BOC$. Suppose $AB<AB$. $BA\perp BA'$ $\Longrightarrow$ $\widehat{CBD}\equiv 2\cdot \widehat{CBA'}$ $\Longrightarrow$ $\widehat{OBD}\equiv 90^{\circ}+C-B$. $\left\{\begin{array}{c}BC=2\rho\sin\widehat{BOC}\\\\ m(\widehat{BOC})=2A\end{array}\right\|\Longrightarrow\left\|\begin{array}{c}\rho =\frac{a}{2\sin 2A}\\\\ a=2R\sin A\\\\ \sin 2A=2\sin A\cos A\end{array}\right\|\Longrightarrow\rho =\frac{R}{2\cos A}\ .$ Therefore, $\left\{\begin{array}{c}OD=2\rho \sin\widehat{OBD}\\\\ \rho=\frac{R}{2\cos A}\\\\ m(\widehat{OBD})=90^{\circ}+C-B\end{array}\right\|\Longrightarrow\boxed{\ OD=\frac{R\cos (B-C)}{\cos A}\ }$ a.s.o.

If we change circumcenter $O$ by incenter $I$ we will get the inequalities $8r^{3}\le ID\cdot OE\cdot IF\le R^{3}$ and we will make a question ourself Find the locus of $P$ in triangle such that $PD\cdot PE\cdot PF\le R^{3}$(or $\ge R^{3}$)