Saudi Arabia Mathematical Competitions

Let $K_A,K_B,K_C$ be the areas of triangle $BPC$, $CPA$, $APB$, respectively. Let $A^{\prime }$ and $P^{\prime }$ be the projections of $A$ and $P$ on side $BC$. Triangles $AL{A}^{\prime }$ and $PL{P}^{\prime }$ are similar, hence we have $$\frac{AL}{PL}=\frac{A{A}^{\prime }}{P{P}^{\prime }}=\frac{A{A}^{\prime }\cdot BC}{P{P}^{\prime }\cdot BC}=\frac{K}{K_A},\text{\hspace{1em}(1)}$$ where $K$ is area of triangle $ABC$. From (1) we get $$\frac{AP}{PL}=\frac{K_B+K_C}{K_A}$$ In analogous way we obtain $$\frac{BP}{PM}=\frac{K_C+K_A}{K_B},\frac{CP}{PN}=\frac{K_A+K_B}{K_C}.\text{\hspace{1em}(2)}$$ The inequality is equivalent to $$\frac{AP}{PL}\cdot \frac{BP}{PM}\cdot \frac{CP}{PN}\ge 8$$ that is $$\left(K_A+K_B\right)\left(K_B+K_C\right)\left(K_C+K_A\right)\ge 8K_A{K}_B{K}_C.\text{\hspace{1em}(3)}$$ Inequality (3) follows by applying three times the inequality $x+y\ge 2\sqrt{xy}$, where $x,y>0$.



We have equality if and only if $K_A=K_B=K_C=\frac{1}{3}K$, hence if and only if $P=G$, the centroid of triangle $ABC$.

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Alternative solution.

We will use so-called Van Aubel relation, that is $$\frac{AP}{PL}=\frac{AN}{NB}+\frac{AM}{MC}.\text{\hspace{1em}(1)}$$ In order to prove (1), we use Menelaos Theorem for triangle $ABL$ and collinear points $C,P,N$, and for triangle $ACL$ and collinear points $B,P,M$. We get $$\frac{CL}{CB}\cdot \frac{NB}{NA}\cdot \frac{PA}{PL}=1\text{ and }\frac{BL}{BC}\cdot \frac{MC}{MA}\cdot \frac{PA}{PL}=1.\text{\hspace{1em}(2)}$$ From (2) we obtain $$\frac{CL}{CB}\cdot \frac{AP}{PL}+\frac{BL}{BC}\cdot \frac{AP}{PL}=\frac{AN}{NB}+\frac{AM}{MC},$$ hence relation (1) since $CL+BL=BC$. Writing the similar relations to (1) for Cevians $BM$ and $CN$, we have $$\frac{BP}{PM}=\frac{BL}{LC}+\frac{BN}{NA}\text{ and }\frac{CP}{PN}=\frac{CL}{LB}+\frac{CM}{MA}.\text{\hspace{1em}(3)}$$ It follows $$\frac{AP}{PL}\cdot \frac{BP}{PM}\cdot \frac{CP}{PN}=\left(\frac{AN}{NB}+\frac{AM}{MC}\right)\left(\frac{BL}{LC}+\frac{BN}{NA}\right)\left(\frac{CL}{LB}+\frac{CM}{MA}\right)$$ and the inequality follows from AM-GM inequality for two numbers.