Let $ x=PA,y=PB,z=PC,a=BC,b=CA,c=AB,$ the inequ become $ ax+by+cz\geq\frac{abc}{R} $ Infact,$ ar_{1}+br_{2}+cr_{3}=\frac{abc}{2R} $.where $ r_{1},r_{2},r_{3} $ be distance from $ P $ to $a,b,c $ and we know $ a(x+r_{1})+b(y+r_{2})+c(z+r_{3})\geq ah_{a}+bh_{b}+ch_{c}=\frac{3abc}{2R} $ so $ ax+by+cz\geq \frac{abc}{R} $ here under $ S_{ABC}=\frac{abc}{4R} $.

just prove $\sum_{cyc}PA*BC\ge\frac{abc}{R}=4S$. notice that $PA*BC\ge 2((APB)+(APC))$ and its two cyclic ineq,adding them yields a soloution.