Prove inequality inside a triangle "Let $ABC$ be a triangle with centroid $G$. A line $PQ$ is drawed in the triangle such that it passes through $G$ and intersects the sides $AB$ and $AC$ in $P$ and $Q$ respectively."
Prove that:
$$\frac {PB \cdot QC}{PA\cdot QA} \le \frac {1}{4}$$ 
I made an aproximate drawing of this problem. Unfortunately i don't know how to start with this kind of problem and i need some hints. Thanks.

 A: Let $PQ$ intersect $BC$ at point $M$ and WLOG let $M$ be closer to $B$ than $C$. Also let $A'$ be the foot of the median from the vertex $A$
Now by Menelaus' Theorem on $\triangle ACA'$ and the line $M-P-Q$ we have:
$$1 = \frac{AG}{GA'} \cdot \frac{A'M}{CM} \cdot \frac{CQ}{QA} \implies \frac{A'M}{CM} \cdot \frac{CQ}{QA} = \frac 12$$
The latter follows as the centroid divides the median into $2:1$ ratio.
Similarly from the Menelau's Theorem on $\triangle ABA'$ and the line $M-P-Q$ we have:
$$1 = \frac{AG}{GA'} \cdot \frac{A'M}{BM} \cdot \frac{BP}{AP} \implies \frac{A'M}{BM} \cdot \frac{BP}{AP} = \frac 12$$
Now multiply these two relations to get:
$$\frac{BP\cdot CQ}{AP \cdot QA} \cdot \frac{A'M^2}{CM \cdot BM} = \frac 14$$
Thus it remains to prove that $\frac{A'M^2}{CM \cdot BM} \ge 1$. This follows as $$CM \cdot BM = \left(A'M - \frac{CB}{2}\right)\left(A'M + \frac{CB}{2}\right) = A'M^2 - \frac{CB^2}{4} \le A'M^2$$
Hence the proof.
A: Notice the following fact，which may be somewhat unusual

$$\frac{PB}{PA}+\frac{QC}{QA}=1$$where $A,B,C,P,Q$ are defined as what you defined.

I won't try to give the detailed proof, because I believe you can readily find it with the help of the figure below.

Thus, by $G_n \leq A_n$, we have $$ \frac{PB}{PA}\cdot\frac{QC}{QA} \leq \frac{1}{4}\left(\frac{PB}{PA}+\frac{QC}{QA}\right)^2=\frac{1}{4}.$$
A: Let $M\in AC$ and $N\in AB$ such that $MN||BC$, and $G\in MN.$
Also, let $\measuredangle MGQ=\theta$.
Thus, in the standard notation $$MG=GN=\frac{1}{3}a$$ and  by law of sines we obtain:
$$\frac{PN}{\sin\theta}=\frac{\frac{1}{3}a}{\sin\beta}$$ or
$$PN=\frac{a\sin\theta}{3\sin\beta}.$$
Similarly, 
$$MQ=\frac{a\sin\theta}{3\sin\gamma}.$$
Id est, we need to prove that
$$\frac{\left(\frac{1}{3}c+\frac{a\sin\theta}{3\sin\beta}\right)\left(\frac{1}{3}b-\frac{a\sin\theta}{3\sin\gamma}\right)}{\left(\frac{2}{3}c-\frac{a\sin\theta}{3\sin\beta}\right)\left(\frac{2}{3}b+\frac{a\sin\theta}{3\sin\gamma}\right)}\leq\frac{1}{4}$$ or
$$\frac{(\sin\gamma\sin\beta+\sin\alpha\sin\theta)(\sin\gamma\sin\beta-\sin\alpha\sin\theta)}{(2\sin\gamma\sin\beta-\sin\alpha\sin\theta)(2\sin\gamma\sin\beta+\sin\alpha\sin\theta)}\leq\frac{1}{4}$$ or
$$\sin^2\alpha\sin^2\theta\geq0.$$
Done!
