Let $ \alpha = \angle{CAP}$, $ \beta = \angle{ABP}$ and $ \gamma = \angle{BCP}$. Apply Ptolemy in $ PBA_1C$ and get $ PA_1\cdot a = PB\cdot CA_1 + PC\cdot BA_1$. From Sine Theorem in $ \triangle ABA_1$ and in $ \triangle {ACA_1}$, we have $ \displaystyle BA_1 = \frac {c\sin(A - \alpha)}{\sin\gamma}$ and $ CA_1 = \frac {b\sin\alpha}{\sin(B - \beta)}$, hence $ \boxed{PA_1 = PB\cdot\displaystyle\frac {b\sin\alpha}{a\sin(B - \beta)} + PC\cdot\frac {c\sin(A - \alpha)}{a\sin\gamma}}$. Let $ S_A = [PBC]$, $ S_B = [PAC]$ and $ S_C = [PAB]$. Then $ \displaystyle\frac {S_B\cdot PB}{S_A\cdot PA} = \frac {b\sin\alpha}{a\sin(B - \beta)}$, hence $ \boxed{PA_1 = \displaystyle\frac {S_BPB^2 + S_CPC^2}{S_APA}}$. Let $ x = S_APA^2$, $ y = S_BPB^2$ and $ z = S_CPC^2$. Then the inequality to prove rewrites as $ \left(1 + \displaystyle\frac {2x}{y + z}\right)\left(1 + \frac {2x}{y + z}\right)\left(1 + \frac {2x}{y + z}\right)\ge8$, which follows immediately after multiplying $ \displaystyle 1 + \frac {2x}{y + z}\ge\frac {2\sqrt {(x + y)(x + z)}}{y + z}$ with its analogues.

Fang-jh wrote: Let $ P$ be an arbitary point inside to $ \triangle ABC$ . Denote the second intersection $ A_1$ of the line $ AP$ with the circumcircle of $ \triangle PBC$ . Define similarly the points $ B_{1}$ , $ C_{1}$ . Prove that $ \prod(1 + 2\cdot\frac {PA}{PA_{1}})\ge 8$ and $ \sum\frac {PA}{PA_1}\ge\frac 32$ . Denote the areas $ [PBC]=x$ , $ [APC]=y$ , $ [ABP]=z$ , the power $ p_w(X)$ of the point $ X$ w.r.t. the circumcircle $ w$ of $ \triangle ABC$ , the point $ D\in AA_1\cap BC$ and the second intersection $ D_1$ of the line $ A_1$ with the circumcircle $ w$ . Lemma. $ \frac {PA}{PA_1}=\frac {x\cdot PA^2}{y\cdot PB^2+z\cdot PC^2}$ a.s.o. Proof I (similarly to Freemind's, without trigonometry). Apply the Ptolemy' theorem to the quadrilateral $ BPCA_1$ : $ PA_1\cdot BC=PB\cdot CA_1+PC\cdot BA_1$ , i.e. $ PA_1=PB\cdot\frac {CA_1}{BC}+PC\cdot\frac {BA_1}{BC}\ \ (1)$ . Therefore, $ \|\begin{array}{ccc} \frac {CA_1}{BC}=\frac {CA_1}{PB}\cdot\frac {PB}{BC}=\frac {DC}{DP}\cdot\frac {PB}{PC}=\frac {PB}{PA}\cdot\frac {PA}{PD}\cdot\frac {CD}{CB}=\frac {PB}{PA}\cdot\frac {[APC]}{[PDC]}\cdot\frac {[PDC]}{[PBC]} & \implies & \frac {CA_1}{BC}=\frac yx\cdot\frac {PB}{PA}\ \ (2)\\\\ \frac {BA_1}{BC}=\frac {BA_1}{PC}\cdot\frac {PC}{BC}=\frac {DB}{DP}\cdot\frac {PC}{BC}=\frac {PC}{PA}\cdot\frac {PA}{PD}\cdot\frac {BD}{BC}=\frac {PC}{PA}\cdot\frac {[ABP]}{[PDB]}\cdot\frac {[PDB]}{[PBC]} & \implies & \frac {BA_1}{BC}=\frac zx\cdot\frac {PC}{PA}\ \ (3)\end{array}$ Using the above relations $ (1)$ , $ (2)$ , $ (3)$ obtain $ PA_1=\frac {y\cdot PB^2+z\cdot PC^2}{x\cdot PA}$ , i.e. $ \boxed {\ \frac {PA}{PA_1}=\frac {x\cdot PA^2}{y\cdot PB^2+z\cdot PC^2}\ }$ . Proof II. Observe that $ AD\cdot DD_1=PD\cdot DA_1$ and $ AD\cdot PD_1=AD\cdot (PD+DD_1)=$ $ AD\cdot PD+AD\cdot DD_1=$ $ AD\cdot PD+PD\cdot DA_1=$ $ PD\cdot AA_1$ . Thus , $ \boxed {\ AD\cdot PD_1=PD\cdot AA_1\ }\ \ (4)$ . I"ll use the remarkable relation $ \boxed {\boxed {\ x\cdot PA^2+y\cdot PB^2+z\cdot PC^2=-(x+y+z)\cdot p_w(P)\ }}$ . Therefore, ${ PA+\frac {y\cdot PB^2+z\cdot PC^2}{x\cdot PA}=\frac {[ABC]}{[PBC]}\cdot \frac {PA\cdot PD_1}{PA}=\frac {AD}{PD}\cdot PD_1}$ . Using the relation $ (4)$ obtain : $ \frac {y\cdot PB^2+z\cdot PC^2}{x\cdot PA}=\frac {AD\cdot PD_1}{PD}-PA=\frac {PD\cdot AA_1}{PD}-PA=AA_1-PA=PA_1$ , i.e. $ PA_1=\frac {y\cdot PB^2+z\cdot PC^2}{x\cdot PA}$ . In conclusion, $ \boxed {\ \frac {PA}{PA_1}=\frac {x\cdot PA^2}{y\cdot PB^2+z\cdot PC^2}\ }$ . Remark. Denote $ u=y\cdot PB^2+z\cdot PC^2$ , $ v=z\cdot PC^2+x\cdot PA^2$ , $ w=x\cdot PA^2+y\cdot PB^2$ . Thus, $ \prod(1+2\cdot\frac {PA}{PA_1})=\prod\frac {v+w}{u}\ge 8$ and $ \sum(1+2\cdot\frac {PA}{PA_1})=\sum\frac {v+w}{u}\ge 6$ , i.e. $ \sum\frac {PA}{PA_1}\ge\frac 32$ .

Another idea is to invert with respect to point $ P$. Then circles $ BCA_{1}$, $ ACB_{1}$, $ ABC_{1}$ all go to straight lines. So if $ T'$ represents image of point $ T$, then the figure simplifies down to cevians $ A'A_{1}'$, $ B'B_{1}'$ and $ C'C_{1}'$ representing cevians of triangle $ A'B'C'$ (with $ A_{1}'$ on $ B'C'$ etc ), that concur at $ P$. Since inversion switches ratios involving inversion centre, (length $ x$ becomes proportional to $ \frac {1}{x}$), our new inequality is that if $ x = \frac {PA_{1}'}{A'P}$, $ y = \frac {PB_{1}'}{B'P}$, $ z = \frac {PC_{1}'}{C'P}$, then $ (1+2x)(1+2y)(1+2z) \geq 8$. This is straightforward with $ AM-GM$ once we have that $ \sum_{cyc} \frac {A'A'_{1}}{PA'_{1}} = 2$, which is simple to prove by looking at area ratios (eg $ \frac { |A'B'C'| }{ |PB'C'}$ etc ).

When does the equality occurs?

Consider the inversion $\mathcal{I} : X \mapsto X'$ with pole $P$ and arbitrary radius $r.$ Because $P, B, C, A_1$ are concyclic, it follows that $B', C', A_1'$ are collinear. Hence, $\overline{A'A_1'}, \overline{B'B_1'}, \overline{C'C_1'}$ are just cevians in $\triangle A'B'C'.$ Furthermore, because $PA \cdot PA' = PA_1 \cdot PA_1' = r^2$, it follows that $PA : PA_1 = PA_1' : PA'.$ Therefore, \[\frac{PA}{PA_1} = \frac{PA_1'}{PA'} = \frac{1}{\frac{PA'}{PA_1'}} = \frac{1}{\frac{AA_1'}{PA_1'} - 1} = \frac{1}{\frac{[A'B'C']}{[PB'C']} - 1} = \frac{[PB'C']}{[A'B'C'] - [PB'C']},\] where we have converted the length ratios into area ratios. Let us denote $x := [PB'C'], y := [PC'A'], z := [PA'B']$ so that $[A'B'C'] = x + y + z.$ It follows from AM-GM that \[\prod \left(1 + 2 \cdot \frac{PA}{PA_1}\right) = \prod \left(1 + \frac{2x}{y + z}\right) = \prod \frac{(x + y) + (x + z)}{y + z} \ge \prod \frac{2\sqrt{(x + y)(x + z)}}{y + z} = 8,\] as desired. $\square$

We employ barycentric coordinates. Let $P=(d,e,f)$ with $d+e+f=1$. Then the equation of the circumcircle of $\triangle{PBC}$ is \[-a^2yz-b^2zx-c^2xy+\frac{a^2ef+b^2fd+c^2de}{d(d+e+f)}\cdot x(x+y+z)=0.\]The point $(t:e:f)$ on this circle besides $t=d$ is $t=-\frac{a^2ef(d+e+f)}{a^2ef+b^2fd+c^2de}$ by Vieta's formula. Then \[\frac{PA}{A_1A}=\frac{e}{d+e+f}\div\frac{e}{t+e+f}=\frac{t+e+f}{d+e+f}=e+f-\frac{a^2ef}{a^2ef+b^2fd+c^2de}.\]Thus by symmetry, \[\frac{PA}{A_1A}+\frac{PB}{B_1B}+\frac{PC}{C_1C}=2(d+e+f)-\frac{a^2ef+b^2fd+c^2de}{a^2ef+b^2fd+c^2de}=1.\]But $\frac{PA}{PA_1}=\frac{PA/A_1A}{1-PA/A_1A}$ so it suffices to prove that $\frac{1+x}{1-x}\cdot\frac{1+y}{1-y}\cdot\frac{1+z}{1-z}\geq8$ when $x+y+z=1$. But this inequality is equivalent to \[9(x+y+z)+9xyz\geq7+7(yz+zx+xy).\]Using Muirhead notation, Muirhead allows us to write \[[3,0,0]+6[2,1,0]+\frac{7}{2}[1,1,1]\geq\frac{7}{2}[1,1,1]+7[2,1,0].\]But this is \[2(x+y+z)^3+9xyz\geq7(x+y+z)(yz+zx+xy),\]from which the desired inequality follows.