Clearly, $\angle PBC=A+B,\angle PCA=B+C,\angle PAB=C+A$. Now applying sine rule on the triangle $PBC$, its easy to get that the circumradius of $\odot PBC=\frac {2R\sin A}{\sin C}$. And $BX=\frac {2R\sin A\sin B}{\sin C}$ and $CX=\frac {2R \sin^{2} A}{\sin C}$. So $[BCX]=\frac {1}{2}BX.CX.\sin \angle BXC=\frac {1}{2}BX.CX.\sin C=\frac {2R^{2}\sin^{3} A\sin B}{\sin C}$. We get similar expressions for $[CAY],[ABZ]$. So summing them we get $\sum_{cyc}2R^{2}\frac {\sin^{3}A\sin B}{\sin C}\geq 6R^{2}\sin A\sin B\sin C=3[ABC]$.(By AM-GM inequality). So done.

i am taking $\angle PAC=\angle PBA=\angle PCB=\omega$(it is the same thing) $\angle PAC+\angle ACP=\angle XPC=\omega+C-\omega=C$ similarly $\angle XPB=A$ note that $BPCX$ is cyclic implies $\angle CBX=C,\angle BCX=A$ $\Longrightarrow \Delta ABC\simeq \Delta CXB$ thus $\frac{[BXC]}{[ABC]}=\frac{BC^2}{AC^2}$ similarly we can do for the rest and apply A.M,G.M